Given : A circle, 2x2+2y2=5 and a parabola y2=45xStatement-1 : An equation of a common tangent to these curves is y=x+5.Statement-2 : If the line, y=mx+5m(m≠0) is their common tangent, then m satisfies m4−3m2+2=0.

Let a tangent to the parabola be y=mx+5m (m≠0)As it is a tangent to the circle x2+y2=5/2, we have (5m)=521+m2⇒(1+m2)m2=2which gives m4+m2−2=0⇒(m2−1)(m2+2)=0As m∈R,m2=1 ∴m=±1Also m=±1 does satisfy m4−3m2+2=0Hence, common tangents are y=x+5 and y=−x−5

Given : A circle, 2x2+2y2=5 and a parabola y2=45xStatement-1 : An equation of a common tangent to these curves is y=x+5.Statement-2 : If the line, y=mx+5m(m≠0) is their common tangent, then m satisfies m4−3m2+2=0.

Let a tangent to the parabola be y=mx+5m (m≠0)As it is a tangent to the circle x2+y2=5/2, we have (5m)=521+m2⇒(1+m2)m2=2which gives m4+m2−2=0⇒(m2−1)(m2+2)=0As m∈R,m2=1 ∴m=±1Also m=±1 does satisfy m4−3m2+2=0Hence, common tangents are y=x+5 and y=−x−5

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Given : A circle, $2 x^{2} + 2 y^{2} = 5$ and a parabola $y^{2} = 4 \sqrt{5} x$
Statement-1 : An equation of a common tangent to these curves is $y = x + \sqrt{5}$ .
Statement-2 : If the line, $y = m x + \frac{\sqrt{5}}{m} (m \neq 0)$ is their common tangent, then m satisfies $m^{4} - 3 m^{2} + 2 = 0$ .

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Statement-1 is false, Statement- 2 is true

Statement-1 is true, Statement-2 is false.

Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.

Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1.

answer is A.

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Detailed Solution

Let a tangent to the parabola be $y = m x + \frac{\sqrt{5}}{m} (m \neq 0)$
As it is a tangent to the circle $x^{2} + y^{2} = 5 / 2$ , we have $(\frac{\sqrt{5}}{m}) = \frac{\sqrt{5}}{\sqrt{2}} \sqrt{1 + m^{2}} \Rightarrow (1 + m^{2}) m^{2} = 2$
which gives $m^{4} + m^{2} - 2 = 0 \Rightarrow (m^{2} - 1) (m^{2} + 2) = 0$
As $m \in R, m^{2} = 1 ∴ m = \pm 1$
Also $m = \pm 1$ does satisfy $m^{4} - 3 m^{2} + 2 = 0$
Hence, common tangents are $y = x + \sqrt{5}$ and $y = - x - \sqrt{5}$