You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Pair of Tangents from an External Point

Suppose that ...

Question

Suppose that the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ are $(f_{1}, 0)$ and $(f_{2}, 0)$ where $f_{1} > 0$ and $f_{2} < 0$ . Let $P_{1}$ and $P_{2}$ be two parabolas with a common vertex at $(0, 0)$ and with foci at $(f_{1}, 0)$ and $(2 f_{2}, 0)$ , respectively. Let $T_{1}$ be a tangent to $P_{1}$ which passes through $(2 f_{2}, 0)$ and $T_{2}$ be a tangent to $P_{2}$ which passes through $(f_{1}, 0)$ . If $m_{1}$ is the slope of $T_{1}$ and $m_{2}$ is the slope of $T_{2}$ , then the value of $(\frac{1}{m_{1}^{2}} + m_{2}^{2})$ is .

Open in App

Solution

Ellipse : $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$
$c^{2} = a^{2} - b^{2} = 9 - 5 \Rightarrow c = 2$
$∴ f_{1} = 2$ and $f_{2} = - 2$
$P_{1} : y^{2} = 8 x P_{2} : y^{2} = - 16 x$

Equation of tangent : $y = m x + \frac{a}{m}$
$T_{1} : y = m_{1} x + \frac{2}{m_{1}}$
Putting point $(2 f_{2}, 0) = (- 4, 0)$
$\Rightarrow 0 = - 4 m_{1} + \frac{2}{m_{1}} \Rightarrow m_{1}^{2} = \frac{1}{2}$

$T_{2} : y = m_{2} x - \frac{4}{m_{2}}$
Putting point $(f_{1}, 0) = (2, 0)$
$\Rightarrow 0 = 2 m_{2} - \frac{4}{m_{2}} \Rightarrow m_{2}^{2} = 2$

Hence $(\frac{1}{m_{1}^{2}} + m_{2}^{2}) = 2 + 2 = 4$

Suggest Corrections

Similar questions

Q. Suppose that the foci of the ellipse

\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1

are

(f_{1}, 0)

and

(f_{2}, 0)

, where

f_{1} > 0

and

f_{2} < 0

let

P_{1}

and

P_{2}

be two parabolas with a common vertex at (0,0) with foci at

(f_{1}, 0)

and

(2 f_{2}, 0)

respectively. Let

T_{1}

be a tangent to

P_{1}

which passes through

(2 f_{2}, 0)

and

T_{2}

be a tangent to

P_{2}

which passes through

(f_{1}, 0)

. If

m_{1}

is the slope of

T_{1}

and

m_{2}

is the slope of

T_{2},

then the value of

(\frac{1}{m_{1}^{2}} + m_{2}^{2})

Suppose that the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ are $(f_{1}, 0)$ and $(f_{2}, 0)$ where $f_{1} > 0$ and $f_{2} < 0$ . Let $P_{1}$ and $P_{2}$ be two parabolas with a common vertex at $(0, 0)$ and with foci at $(f_{1}, 0)$ and $(2 f_{2}, 0)$ respectively. Let $T_{1}$ be a tangent to $P_{1}$ which passes through $(2 f_{2}, 0)$ and $T_{2}$ be a tangent to $P_{2}$ which passes through $(f_{1}, 0)$ . If $m_{1}$ is the slope of $T_{1}$ and $m_{2}$ is the slope of $T_{2}$ , then the value of $(\frac{1}{m_{1}^{2}} + m_{2}^{2})$ is

In $R^{3}$ , consider the planes $P_{1} : y = 0$ and $P_{2} : x + z = 1$ . Let $P_{3}$ be a plane , different from $P_{1}$ and $P_{2}$ , which passes through the intersection of $P_{1}$ and $P_{2}$ . If the distance of the point (0, 1, 0) from
$P_{3}$ is 2, then which of the following relation(s) is/are true?

Q. Slope of common tangents of parabola

(x - 1)^{2} = 4 (y - 2)

and ellipse

(x - 1)^{2} + \frac{(y - 2)^{2}}{2} = 1

are

m_{1}

and

m_{2}

then

m_{1}^{2} + m_{2}^{2}

is equal to

Two bodies of masses m₁ and m₂ having equal kinetic energies. If p₁ and p₂ are their momentum respectively, Then the ratio of p₁ and p₂ is.

Join BYJU'S Learning Program

Ellipse : x29+y25=1 c2=a2−b2=9−5⇒c=2 ∴f1=2 and f2=−2 P1:y2=8xP2:y2=−16x Equation of tangent :y=mx+am T1:y=m1x+2m1 Putting point (2f2,0)=(−4,0) ⇒0=−4m1+2m1⇒m21=12 T2:y=m2x−4m2 Putting point (f1,0)=(2,0) ⇒0=2m2−4m2⇒m22=2 Hence (1m21+m22)=2+2=4