Test Index

Parabola Part 1

Section: Mathematics

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Question : 45 of 100

Marks: +1, -0

Let

L

be a tangent line to the parabola

y^{2}=4 x-20

(6,2)

L

is also a tangent to the ellipse

\;\frac{x^{2}}{2}+\;\frac{y}{b}=1

, then the value of

b

[17 Mar 2021 Shift 2]

11
14
16
20

Solution:

Given, equation of parabola is

y^{2}=4 x-20

....(i) Differentiating

Eq. (i) w.r.t.

x

, we get

2 y \;\frac{dy}{dx}=4 \Rightarrow \;\frac{dy}{dx}=\;\frac{2}{y}

\therefore \;\; \left(\;\frac{dy}{dx}\right)_{(6,2)}=

Slope of tangent at

(6,2)=\;\frac{2}{2}=1

\therefore

Equation of tangent is

y-2=1(x-6) \Rightarrow x-y-4=0

...(ii)

As, we know the condition of tangency to the ellipse,

A straight line

y=m x+c

will be tangent to the ellipse

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

iff,

c^{2}=a^{2} m^{2}+b^{2}

\therefore 16=2(1)^{2}+b

[ Here,

c=-4, m=1, b=\sqrt{b}, a=1

]

⇒ b=14

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