Parabola Ques 32
- Given A circle, $2 x^{2}+2 y^{2}=5$ and a parabola, $y^{2}=4 \sqrt{5} x$.
Statement I An equation of a common tangent to these curves is $y=x+\sqrt{5}$.
Statement II If the line, $y=m x+\frac{\sqrt{5}}{m}(m \neq 0)$ is the common tangent, then $m$ satisfies $m^{4}-3 m^{2}+2=0$.
(2013 Main)
(a) Statement I is correct, Statement II is correct, Statement II is a correct explanation for Statement I
(b) Statement I is correct, Statement II is correct, Statement II is not a correct explanation for Statement I
(c) Statement I is correct, Statement II is incorrect
(d) Statement I is correct, Statement II is incorrect
Objective Questions II
(One or more than correct option)
Show Answer
Solution:
- Equation of circle can be rewritten as $x^{2}+y^{2}=\frac{5}{4}$.
$$ \text { Centre } \rightarrow(0,0) \text { and radius } \rightarrow \sqrt{\frac{5}{2}} $$
Let a common tangent be
$$ y=m x+\frac{\sqrt{5}}{m} \Rightarrow m^{2} x-m y+\sqrt{5}=0 $$
The perpendicular from centre to the tangent is equal to radius of the circle.
$$ \begin{array}{ll} \therefore & \frac{\sqrt{5} / m}{\sqrt{1+m^{2}}}=\sqrt{\frac{5}{2}} \\ \Rightarrow & m \sqrt{1+m^{2}}=\sqrt{2} \\ \Rightarrow & m^{2}\left(1+m^{2}\right)=2 \\ \Rightarrow & m^{4}+m^{2}-2=0 \\ \Rightarrow & \left(m^{2}+2\right)\left(m^{2}-1\right)=0 \\ \Rightarrow & m= \pm 1 \quad\left[\because m^{2}+2 \neq 0, \text { as } m \in R\right] \end{array} $$
$\therefore y= \pm(x+\sqrt{5})$, both statements are correct as $m \pm 1$ satisfies the given equation of Statement II.
BETA
