Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Parabola Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2B Parabola Important Questions

Question 1.
Find the equation of the parabola whose focus is S (1, -7) and vertex is A(1, -2). [T.S. Mar. 15]
Solution:
Let S = (1, -7), A(1, -2)
h = 1, k = -2, a = -2 + 7 = 5
The Axis of the parabola is parallel to the y-axis
The equation of the parabola is
(x – h)2 = – 4a (y – k)
(x – 1)2 = – 20(y + 2)
x2 – 2x + 1, = – 20y – 40
⇒ x2 – 2x + 20y + 41 – 0. Question 2.
If the normal at the point t1 on the parabola y2 =, 4ax meets it again at point t2 then prove that t1t2 + t12 + 2 = 0. [May 07]
Solution:
Equation of normal is
y – y1 = $$\frac{-y_{1}}{2 a}$$ (x – x1)
y – 2at1 = $$\frac{-2 a t_{1}}{2 a}$$ (x – at12) ……………… (i)
Equation of the line (i) again meets parabola at (at22, 2at1)
∴ 2at2 – 2at1 = t1 (at22 – at12)
$$-\frac{2}{t_{1}}$$ = t1 + t2 ⇒ -2 = t12 + t1t2
⇒ t12 + t1t2 + 2 = 0

Question 3.
Find the coordinates of the vertex and focus, and the equations of the directrix and axes of the y – x + 4y + 5 = 0. [Mar 05]
Solution:
y2 – x + 4y + 5 = 0 ⇒ (y – (-2))2 = (x – 1), comparing with (y – k)2 = 4a(x – h),we get (h, k) = (1, -2) and a = $$\frac{1}{4}$$, coordinates of the vertex (h, k) = (1 ,-2)
coordinates of the focus (h + a, k) = ($$\frac{5}{4}$$, -2)
Equation of the directrix x – h + a = 0
i.e., 4x – 3 = 0
Equation of the axis y – k = 0.
i.e., y + 2 = 0 Question 4.
Find the coordinates of the points on the parabola y2 = 2x whose focal distance is $$\frac{5}{2}$$. [A.P. Mar. 15, Mar. 13]
Solution:
Let P(x1, y1) be a point on the parabola
y2 = 2x whose focal distance is $$\frac{5}{2}$$ then
y12 = 2x1 and x1 + a = $$\frac{5}{2}$$
⇒ x1 + $$\frac{1}{2}$$ = $$\frac{5}{2}$$ ⇒ x1 = 2
∴ y12 = 2(2) = 4 ⇒ y1 = ±2
∴ The required points are (2, 2) and (2, -2)

Question 5.
Find the value of k if the line 2y = 5x + k is a tangent to the parabola y2 = 6x. [T.S. Mar. 16]
Solution:
Given line is 2y = 5x + k
⇒ y = ($$\frac{5}{2}$$)x + ($$\frac{k}{2}$$)
Comparing y = ($$\frac{5}{2}$$)x + ($$\frac{k}{2}$$) with y = mx + c
We get m = $$\frac{5}{2}$$, c = $$\frac{k}{2}$$
y = ($$\frac{5}{2}$$)x + $$\frac{k}{2}$$ is a tangent to y2 = 6x
c = $$\frac{a}{m}$$
⇒ $$\frac{k}{2}=\frac{\left(\frac{3}{2}\right)}{\left(\frac{5}{2}\right)}$$ ⇒ k = $$\frac{6}{5}$$

Question 6.
Show that the equations of common tangents to the circle x2 + y2 = 2a2 and the parabola y2 = 8ax are y = ± (x + 2a). [Mar. 06]
Solution:
The equation of tangent to parabola
y2 = 8ax is y = mx + $$\frac{2a}{m}$$
m2x – my + 2a = 0 ……………….. (1)
If (i) touches circle x2 + y2 = 2a2, then the length of perpendicular from its centre (0, 0) to (i) must be equal to the radius a$$\sqrt{2}$$ of the circle.
$$\left|\frac{2 a}{\sqrt{m^{2}+m^{4}}}\right|=a \sqrt{2}$$
or 4 = 2 (m4 + m2)
m4 + m2 – 2 = 0
(m2 + 2) (m2 – 1) = 0 or m = ±1
Required tangents are
y = (1) x + $$\frac{2 a}{(1)}$$, y = (-1) x + $$\frac{2 a}{(-1)}$$
y = ± (x + 2a) Question 7.
Find the co-ordinates of the point on the parabola y2 = 8x whose focal distance is 10. [T.S. Mar. 17] [A.P. Mar. 16; Mar. 14, 11]
Solution:
Equation of the parabola is
y2 = 8x
4a = 8 ⇒ a – 2 Co-ordinates of the focus S are (2, 0) Suppose P(x, y) is the point on the parabola.
Given SP = 10 ⇒ SP2 = 100
(x – 2)2 + y2 = 100
But y2 = 8x
⇒ (x – 2)2 + 8x = 100
⇒ x2 – 4x + 4 + 8x – 100 = 0
⇒ x2 + 4x – 96 = 0 ⇒ (x + 12) (x – 8) = 0
x + 12 = 0 or x – 8 = 0
x = -12, or 8
Case (i) x = 8
y2 = 8.x = 8.8 = 64
y = ±8
Co-ordinates of the required points are (8, 8) and (8, -8)
Case (ii) x = -12
y2 = 8(-12) = -96 < 0
y is not real.

Question 8.
If ($$\frac{1}{2}$$, 2) is one extermity of a focal chord of the parabola y2 = 8x. Find the co-ordinates of the other extremity. [May 06]
Solution:
A = ($$\frac{1}{2}$$, 2); S = (2, 0)
B = (x1, y1) ⇒ ($$\frac{y_{1}^{2}}{8}$$, y1)
ASB is a focal chord.
∴ Slopes of SA and BS are same. or 4y12 + 24y1 – 64 = 0
⇒ y12 + 6y1 – 16 = 0
⇒ (y1 + 8) (y, – 2) = 0
y1 = 2, 8
x1 = $$\frac{1}{2}$$, 8; So (8, -8) other extremity. Question 9.
Show that the common tangent to the parabola y2 = 4ax and x2 = 4by is xa1/3 + yb1/3 + a2/3b2/3 = 0 [T.S. Mar. 17] [A.P. Mar. 16]
Solution:
The equations of the parabolas are
y2 = 4ax ………………. (1)
and x2 = 4by ……………… (2)
Equation of any tangent to (1) is of the form
y = mx + $$\frac{a}{m}$$ …………….. (3)
If the line (3) is a tangent to (2) also, we must get only one point of intersection of (2) and (3).
Substituting the value of y from (3) in (2), we get x2 = 4b (mx + $$\frac{a}{m}$$) is 3x2 – 4bm2x – 4ab = 0 should have equal roots Therefore its discriminent must be zero. Hence
16b2m4 – 4m (-4ab) = 0
16b(bm4 + am) = 0
m(bm3 + a) = 0 But m ≠ 0
∴ m = – a1/3/b1/3 substituting in (3) the equation of the common tangent becomes
y = $$-\left(\frac{a}{b}\right)^{1 / 3} x+\frac{a}{\left(-\frac{a}{b}\right)^{1 / 3}}$$ or
a1/3x + b1/3y + a2/3b2/3 = 0 .

Question 10.
Prove that the area of the triangle formed by the .tangents at (x1, y1), (x2, y2) and (x3, y3) to the parabola y2 = 4ax (a > 0) is $$\frac{1}{16a}$$|(y1 – y2) (y2 – y3) (y3 – y1)| sq. units. [T.S. Mar. 15]
Solution:
Let D(x1, y1) = (at12, 2at1),
E(x2, y2) = (at22, 2at2),
and F(x3, y3) = (at32, 2at3), be three points on the parabola
y2 = 4ax (a > 0).
The equation of the tangents at D, E and F are
t1y = x + at12 ……………… (1)
t2y = x + at22 ………………. (2)
t3y = x.+ at32 ………………. (3)
(1) – (2) ⇒ (t1 – t2) y = a(t1 – t2) (t1 + t2)
⇒ y = a(t1 + t2) substituting in (1)
we get x = at1t2.
∴ The point of intersection of the tangents at D and E is say P(at1t2, a(t1 + t2))
Similarly the points of intersection of tangent at E, F and at F, D are Q(at2t3, a(t2 + t3) and R(at3t1 a(t3 + t1)) respectively
Area of ∆PQR $$\frac{1}{16a}$$ |2a(t1 – t2) 2a(t2 – t3) 2a(t3 – t1)|
$$\frac{1}{16a}$$|(y1 – y2) (y2 – y3) (y3 – y1)| sq. units.

Question 11.
Prove that the two parabolas y2 = 4ax and x2 = 4by intersect (other than the origin) at an angle of Tan-1 $$\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]$$ [Mar. 14]
Solution:
Without loss of generality we assume a > 0 and b > 0.
Let P(x, y) be the point of intersection of the parabolas other than the origin. Then
y4 = 16a2x2
= 16a2(4by)
= 64a2by
∴ y[y3 – 64a2b] = 0
=> y3 – 64a2b = 0
=> y = (64a2b)1/3 [∵ y > 0]
= 4a2/3b1/3 Also from y2 = 4ax, x = $$\frac{16 a^{4 / 3} b^{2 / 3}}{4 a}$$
= 4a1/3b2/3
∴ P = (4a1/3b2/3, 4a2/3b1/3)
Differentiating both sides of y2 4ax w.r.t ‘x’, we get
$$\frac{d y}{d x}=\frac{2 a}{y}$$
∴ $$\left[\frac{\mathrm{dy}}{\cdot \mathrm{dx}}\right]_{\mathrm{p}}=\frac{2 \mathrm{a}}{4 \mathrm{a}^{2 / 3} \mathrm{~b}^{1 / 3}}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{1 / 3}$$
If m1 be the slope of the tangent at P to y2 = 4ax, then
m1 = $$\frac{1}{2}\left(\frac{a}{b}\right)^{1 / 3}$$
Similarly, we get m2 = $$2\left(\frac{a}{b}\right)^{1 / 3}$$ where m2 is the slope of the tangent at P to x2 = 4by.
If θ is the acute angle between the tangents to the curves at P, then
tan θ = $$\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}$$
so that θ = $$\tan ^{-1}\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]$$ Question 12.
Find the coordinates of the vertex and focus, and the equations of the directrix and axes of the following parabolas.
(i) y2 = 16x
(ii) x2 = -4y
(iii) 3x2 – 9x + 5y – 2 = 0
(iv) y2 – x + 4y + 5 = 0 [Mar. 05]
Solution:
i) y2 = 16x, comparing with y2 = 4ax,
we get 4a = 16 ⇒ a = 4
The coordinates of the vertex = (0, 0)
The coordinates of the focus = (a, 0) = (4, 0)
Equation of the directrix: x + a = i.e., x + 4 = 0
Axis of the parabola y = 0

ii) x2 = -4y, comparing with x2 = -4ay,
we get 4a = 4 ⇒ a = 1
The coordinates of the vertex = (0, 0
The coordinates of the focus = (0, -a) = (0, 1)
The equation of the directrix y – a = 0
i.e., y – 1 = 0 .
Equation of the axis x = 0

iii) 3x2 – 9x + 5y – 2 = 0
3(x2 – 3x) = 2 – 5y
⇒ 3(x2 – 2x($$\frac{3}{2}$$) + $$\frac{9}{4}$$) = 2 – 5y + $$\frac{27}{4}$$
(x – $$\frac{3}{2}$$)2 = –$$\frac{5}{3}$$ (y – $$\frac{7}{4}$$),
Comparing with (x – h)2 = -4a (y – k) we get
a = $$\frac{5}{12}$$, h = $$\frac{3}{2}$$, k = $$\frac{7}{4}$$
∴ Coordinates of the vertex = (h, k)
= ($$\frac{3}{2}$$, $$\frac{7}{4 }$$)
Coordinates of the focus = (h, k – a)
= ($$\frac{3}{2}$$, $$\frac{7}{4}$$ – $$\frac{5}{12}$$) = ($$\frac{3}{2}$$, $$\frac{4}{3}$$)
Equation of the directrix is y – k – a = 0
i.e., 6y – 13 = 0
Equation of the axis is x – h = 0
i.e., 2x – 3 = 0

iv) y2 – x + 4y + 5 = 0 ⇒ (y- (-2))2 = (x – 1),
comparing with (y – k)2 = 4a(x – h),we get
(h, k) = (1, -2) and a = $$\frac{1}{4}$$, coordinates of the vertex (h, k) = (1 ,-2)
coordinates of the focus
(h + a, k) = ($$\frac{5}{4}$$, -2)
Equation of the directrix x – h + a = 0
i.e., 4x – 3 = 0
Equation of the axis y – k = 0.
i.e., y + 2 = 0 Question 13.
Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1).
Solution:
The abcissae of the vertex and focus are equal to 3. Hence the axis of the parabola is x = 3, a line parallel to y-axis, focus is above the vertex.
a = distance between focus and vertex = 3.
∴ Equation of the parabola
(x – 3)2 = 4(3) (y + 2)
i.e., (x – 3)2 = 12(y + 2).

Question 14.
Find the coordinates of the points on the parabola y2 = 2x whose focal distance is $$\frac{5}{2}$$. [A.P. Mar. 15, Mar. 13]
Solution:
Let P(x1, y1) be a point on the parabola
y2 = 2x whose focal distance is $$\frac{5}{2}$$ then
y12 = 2x1 and x1 + a = $$\frac{5}{2}$$
⇒ x1 + $$\frac{1}{2}$$ = $$\frac{5}{2}$$ ⇒ x1 = 2
∴ y12 = 2(2) = 4 ⇒ y1 = ±2
∴ The required points are (2, 2) and (2, -2)

Question 15.
Find the equation of the parabola passing through the points (-1, 2), (1, -1) and (2, 1) and having its axis parallel to the X-axis.
Solution:
Since the axis is parallel to x-axis the equation of the parabola is in the form of
x = -ly2 + my + n. .
Since the parabola passes through (-1, 2), we have
-1 = l( 2)2 + m(2) + n
⇒ 4l + 2m + n = – 1 ……………………. (1)
Similarly, since the parabola passes through (1,-1) and (2, 1) we have
l – m + n = 1 ……………… (2)
l + m + n = 2 ……………… (3)
Solving (1), (2) and (3)
we get l = –$$\frac{7}{6}$$, m = $$\frac{1}{2}$$ and n = $$\frac{8}{3}$$.
Hence the equation of the parabola is
x = –$$\frac{7}{6}$$ y2 + $$\frac{1}{2}$$ y + $$\frac{8}{3}$$ (or)
7y2 – 3y + 6x- 16 = 0. Question 16.
A double ordinate of the curve y2 = 4ax is of length 8a. Prove that the line from the vertex to its ends are at right angles.
Solution:
Let P = (at2, 2at) and P’ = (at2, -2at) be the ends of double ordinate PP’. Then
8a = PP’ = $$\sqrt{0+(4 a t)^{2}}$$ = 4at ⇒ t = 2.
∴ P = (4a, 4a), P’= (4a, -4a) .
Slope of $$\overline{\mathrm{AP}}$$ × slope of $$\overline{\mathrm{AP}^{\prime}}$$
= ($$\frac{4a}{4a}$$)(-$$\frac{4a}{4a}$$) = -1
∴ ∠PAP’ = $$\frac{\pi}{2}$$

Question 17.
i) If the coordinates of the ends of a focal chord of the parabola y2 = 4ax are (x1, y1) and (x2, y2), then prove that x1x2 = a2, y1y2 = -4a2.
ii) For a focal chord PQ of the parabola y2 = 4ax, if SP = l and SQ = l’ then prove that $$\frac{1}{l}$$ + $$\frac{1}{l}$$ = $$\frac{1}{a}$$.
Solution:
i) Let P(x1, y1) = (at12, 2at1) and Q(x2, y2) = (at22, 2at1) be two end points of a focal chord.
P, S, Q are collinear
Slope of $$\overline{\mathrm{PS}}$$ = Slope of $$\overline{\mathrm{QS}}$$
$$\frac{2 a t_{1}}{a t_{1}^{2}-a}=\frac{2 a t_{2}}{a t_{2}^{2}-a}$$
t1t22 – t1 = t2t12 – t2
t1t2 (t2 – t1) + (t2 – t1) = 0
1 + t1t2 = 0 ⇒ t1t2 = -1 ………………… (1)
From (1) x1x2 = at12 at22 = a2(t2t1)2 = a2
y1y2 = 2at12at2 = 4a2(t2 t1) = -4a2

ii) Let P(at12, 2at1) and Q(at22, 2at1) be the extremities of a focal chord of the parabola, then t1t2 = -1 (from (1)) Question 18.
If Q is the foot of the perpendicular from a point P on the parabola y2 = 8(x – 3) to its directrix. S is the focus of the parabola and if SPQ is an equilateral triangle then find the length of side of the triangle.
Solution:
Given parabola y2 = 8(x – 3) then
its vertex A = (3, 0) and focus = (5, 0)
[4a = 8 ⇒ a = 2] since PQS is an equilateral triangle Hence length of each side of triangle is ‘8’. Question 19.
The cable of a uniformely loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 72 mt. long is supported by vertical wires attached to the cable, the longest being 30 mts. and the shortest being 6 mts. Find the length of the supporting wire attached to the road-way 18 mts. from the middle. Solution:
Let AOB be the cable [0 is its lowest point and A, B are the highest points]. Let PRQ be the bridge suspended with PR = RQ = 36 mts. (see above Fig.).

PA = QB = 30 mts (longest vertical sup-porting wires)

OR = 6 mts (shortest vertical supporting wire) [the lowest point of the cable is up¬right the mid-point R of the bridge]

Therefore, PR = RQ = 36 mts. We take the origin of coordinates at 0, X-axis along the tangent at O to the cable and the Y-axis along $$\overleftrightarrow{\mathrm{RO}}$$. The equation of the cable would, therefore, be x2 = 4ay for some a > 0. We get B = (36, 24) and 362 = 4a × 24.
Therefore, 4a = $$\frac{36 \times 36}{24}$$ = 54 mts.
If RS = 18 mts. and SC is the vertical through S meeting the cable at C and the X-axis at D, then SC is the length of the supporting wire required. If SC = l mts, then
DC = (l – 6) mts. As such C = (18, l – 6).
Since C is on the cable, 182 = 4a (l – 6)
⇒ l – 6 = $$\frac{18^{2}}{4 a}$$ = $$\frac{18 \times 18}{54}$$ = 6
⇒ l = 12

Question 20.
Find the condition for the straight line lx + my + n = 0 to be a tangent to the parabola y2 = 4ax and find the co-ordinates of the point of contact.
Solution:
Let the line lx + my + n = 0 be a tangent to the parabola y2 = 4ax at (at2, 2at). Then the equation of the tangent at P(t) is x – yt + at2 = 0 then it represents the given line
lx + my + n = 0, then  Question 21.
Show that the straight line 7x + 6y = 13 is a tangent to the parabola y2 – 7x – 8y + 14 = 0 and find the point of contact.
Solution:
Equation of the given line is 7x + 6y = 13, equation of the given parabola is y2 – 7x – 8y + 14 = 0.
By eliminating x, we get the ordinates of the points of intersection of line and parabola adding the equations y2 – 2y + 1 = 0.
i.e., (y – 1 )2 = 0 ⇒ y = 1, 1.
∴ The given line is tangent to the given parabola.
If y = 1 then x = 1 hence the point of contact is (1, 1).

Question 22.
Prove that the normal chord at the point other than origin whose ordinate is equal to its abscissa subtends a right angle at the focus.
Solution:
Let the equation of the parabola be
y2 = 4ax and P(at2, 2at) be any point …………………. (1)
On the parabola for which the abscissa is equal to the ordinate.
i.e., at2 = 2at ⇒ t = 0
or t = 2. But t ≠ 0. Hence the point (4a, 4a) at which the normal is
y + 2x = 2a(2) + a(2)3 (or)
y = (12a – 2x) ………………….. (2)
Substituting the value of
y = 12a – 2x in (1) we get
(12a – 2x)2 = 4ax (or)
x2 – 13ax + 36a2 = (x – 4a) (x – 9a) = 0
⇒ x = 4a, 9a
corresponding values of y are 4a and -6a. Hence the other points of intersection of that normal at P(4a, 4a) to the given parabola is Q(9a, -6a), we have S(a, 0).
Slope of the $$\overline{\mathrm{SP}}$$ = m1 = $$\frac{4a-0}{4a-a}$$ = $$\frac{4}{3}$$,
Slope of the $$\overline{\mathrm{SQ}}$$ = m2 = $$\frac{-6a-0}{9a-a}$$ = $$\frac{3}{4}$$
clearly m1m2 = -1, so that $$\overline{\mathrm{SP}}$$ ⊥ $$\overline{\mathrm{SQ}}$$ Question 23.
From an external point P, tangent are drawn to the parabola y2 = 4ax and these tangent make angles θ1, θ2 with its axis, such that tan θ1 + tan θ2 is constant b. then show that P lies on the line y = bx.
Solution:
Let the coordinates of P be (x1, y1) and the equation of the parabola y2 = 4ax. Any tangent to .the parabola is y = mx + $$\frac{a}{m}$$, if this passes through (x1, y1) then
y1 = mx, + $$\frac{a}{m}$$
i.e., m2x1 – my1 + a = 0 ………………… (1)
Let the roots of (1) be m1, m21.
Then m1 + m2 = $$\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}$$
⇒ tan θ1 + tan θ2 = $$\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}$$
[∵ The tangents make angles θ1, θ2 with its axis (x – axis) then their slopes m1 = tan θ1 and m2 = tan θ2].
∴ b = $$\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}$$ ⇒ y1 = bx1
∴ P(x1, y1) lies on the line y = bx.

Question 24.
Show that the common tangent to the parabola y2 = 4ax and x2 = 4by is xa1/3 + yb1/3 + a2/3b2/3 = 0. [A.P. Mar. 16]
Solution:
The equations of the parabolas are
y2 = 4ax ………………. (1)
and x2 = 4by …………………. (2)
Equation of any tangent to (1) is of the form
y = mx + $$\frac{a}{m}$$
If the line (3) is a tangent to'(2) also, we must get only one point of intersection of (2) and (3).
Substituting the value of y from (3) in (2),
we get x2 = 4b (mx + $$\frac{a}{m}$$) is mx2 – 4bm2x – 4ab = 0 should have equal roots therefore its discriminent must be zero. Hence
16b2m4 – 4m (-4ab) = 0
16b(bm4 + am) = 0
m(bm3 + a) = 0 But m ≠ 0
∴ m = – a1/3/b1/3 substituting in (3) the equation of the common tangent becomes
y = $$-\left(\frac{a}{b}\right)^{1 / 3} x+\frac{a}{\left(-\frac{a}{b}\right)^{1 / 3}}$$ (or)
∴ a1/3 x + b1/3y + a2/3b2/3 = 0 Question 25.
Prove that the area of the triangle formed ‘ by the tangents at (x1, y1), (x2, y2) and (x3, y3) to the parabola y2 = 4ax (a > 0) is $$\frac{1}{16a}$$|(y1 – y2) (y2 – y3) (y3 – y1)| sq. units. [T.S. Mar. 15]
Solution:
Let D(x1, y1) = (at12, 2at1)
E(x2, y2) = (at32, 2at2)
and F(x3, y3) = (at23, 2at3)
be three points on the parabola
y2 = 4ax (a > 0).
The equation of the tangents at D, E and F are
t1y = x + at12 ………………. (1)
t2y = x + at22 ………………. (2)
t3y = x + at32 ………………. (3)
(1) – (2) ⇒ (t1 – t2) y = a(t1 – t2) (t1 + t2)
⇒ y = a(t1 + t2) substituting in (1)
we get x = at1t2
∴ The point of intersection of the tangents at D and E is say P(at1t2, a(t1 + t2))
Similarly the points of intersection of tangent at E, F and at F, D are Q(at2t3, a(t2 + t3) and R(at3t1, a(t3 + t1)) respectively
Area of ∆PQR  Question 26.
Prove that the two parabolas y2 = 4ax and x2 = 4by intersect (other than the origin) at an angle of Tan-1 $$\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]$$ [Mar. 14]
Solution:
Without loss of generality we assume a > 0 and b > 0.
Let P(x, y) be the point of intersection of the parabolas other than the origin. Then
y4 = 16a2x2
= 16a2(4by)
= 64a2by
∴ y[y3 – 64a2b] = 0
⇒ y3 – 64a2b = 0
⇒ y = (64a2b)1/3 [∵ y > 0]
= 4a2/3b1/3 Also from y2 = 4ax, x = $$\frac{16 a^{4 / 3} b^{2 / 3}}{4 a}$$
= 4a1/3b2/3
∴ P = (4a1/3b2/3, 4a2/3b1/3)
Differentiating both sides of y2 4ax w.r.t ‘x’, we get
$$\frac{d y}{d x}=\frac{2 a}{y}$$
∴ $$\left[\frac{\mathrm{dy}}{\cdot \mathrm{dx}}\right]_{\mathrm{p}}=\frac{2 \mathrm{a}}{4 \mathrm{a}^{2 / 3} \mathrm{~b}^{1 / 3}}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{1 / 3}$$
If m1 be the slope of the tangent at P to y2 = 4ax, then
m1 = $$\frac{1}{2}\left(\frac{a}{b}\right)^{1 / 3}$$
Similarly, we get m2 = $$2\left(\frac{a}{b}\right)^{1 / 3}$$ where m2 is the slope of the tangent at P to x2 = 4by.
If θ is the acute angle between the tangents to the curves at P, then
tan θ = $$\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}$$
so that θ = $$\tan ^{-1}\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]$$ Question 27.
Prove that the orthocenter of the triangle formed by any three tangents to a parabola lies on the directrix of the parabola.
Solution:
Let y2 = 4ax be the parabola and
A = (at12, 2at1),
B = (at22, 2at2),
C = (at32, 2at3) be any three points on it.
Now we consider the triangle PQR formed by the tangents to the parabola at A, B, C
where P = (at1t2, a(t1 + t2)),
Q = (at2t3, a(t2 + t3)) and R = (at3t1, a(t3 + t1)).
Equation of $$\overleftrightarrow{\mathrm{QR}}$$ (i.e., the tangent at C) is x – t3 y + at32 = 0.
Therefore, the attitude through P of triangle PQR is
t3x + y = at1t2t3 + a(t1 + t2) ………………….. (1)
Similarly, the attitude through Q is
t1x + y = at1t2t3 + a(t2 + t3) ………………….. (2)
Solving (1) and (2), we get (t3 – t1)
x’ = a(t1 – t3) i.e., x = – a.
Therefore, the orthocenter of the triangle PQR, with abscissa as -a, lies on the directrix of the parabola.