Parabola 2 Question 1

1. If the line $a x+y=c$, touches both the curves $x^{2}+y^{2}=1$ and $y^{2}=4 \sqrt{2} x$, then $|c|$ is equal to

(2019 Main, 10 April, II)

(a) $\frac{1}{\sqrt{2}}$

(b) 2

(c) $\sqrt{2}$

(d) $\frac{1}{2}$

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Answer:

Correct Answer: 1. (c)

Solution:

Key Idea Use the equation of tangent of slope ’ $m$ ’ to the parabola $y^{2}=4 a x$ is $y=m x+\frac{a}{m}$ and a line $a x+b y+c=0$ touches the circle $x^{2}+y^{2}=r^{2}$, if $\frac{|c|}{\sqrt{a^{2}+b^{2}}}=r$.

Since, equation of given parabola is $y^{2}=4 \sqrt{2} x$ and equation of tangent line is $a x+y=c$ or $y=-a x+c$,

then $c=\frac{\sqrt{2}}{m}=\frac{\sqrt{2}}{-a} \quad[\because m=$ slope of line $=-a]$

$[\because$ line $y=m x+c$ touches the parabola

$y^{2}=4 a x$ iff $c=a / m$ ]

Then, equation of tangent line becomes

$$ y=-a x-\frac{\sqrt{2}}{a} $$

$\because$ Line (i) is also tangent to the circle $x^{2}+y^{2}=1$.

$\therefore$ Radius $=1=\frac{\left|-\frac{\sqrt{2}}{a}\right|}{\sqrt{1+a^{2}}} \Rightarrow \sqrt{1+a^{2}}=\left|-\frac{\sqrt{2}}{a}\right|$

$$ \Rightarrow \quad 1+a^{2}=\frac{2}{a^{2}} \quad \text { [squaring both sides] } $$

$\Rightarrow \quad a^{4}+a^{2}-2=0 \Rightarrow\left(a^{2}+2\right)\left(a^{2}-1\right)=0$

$\Rightarrow \quad a^{2}=1 \quad\left[\because a^{2}>0, \forall a \in R\right]$

$\therefore \quad|c|=\frac{\sqrt{2}}{|a|}=\sqrt{2}$



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