Parabola 2 Question 14

14. 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 incorrect, Statement II is correct

Objective Questions II

(One or more than correct option)

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Solution:

  1. Equation of circle can be rewritten as $x^{2}+y^{2}=\frac{5}{2}$.

$$ \text { Centre } \rightarrow(0,0) \text { and radius } \rightarrow \sqrt{\frac{5}{2}} $$

Let 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.



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