Parabola Ques 27
- 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: 27.(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 $=1=\frac{\left|-\frac{\sqrt{2}}{a}\right|}{\sqrt{1+a^{2}}} \Rightarrow \sqrt{1+a^{2}}=\left|-\frac{\sqrt{2}}{a}\right|$
$$ \Rightarrow \quad 1+a^{2}=\frac{2}{a^{2}} \quad \text { [squaring both sides] } $$
$\Rightarrow \quad a^{4}+a^{2}-2=0 \Rightarrow\left(a^{2}+2\right)\left(a^{2}-1\right)=0$
$\Rightarrow \quad a^{2}=1 \quad\left[\because a^{2}>0, \forall a \in R\right]$
$\therefore \quad|c|=\frac{\sqrt{2}}{|a|}=\sqrt{2}$
BETA
