Question:
The line $2x+\sqrt{6y}=2$ is a tangent to the curve $x^{2}-2y^{2}=4$. The point of contact is
The line $2x+\sqrt{6y}=2$ is a tangent to the curve $x^{2}-2y^{2}=4$. The point of contact is
Updated On: Jun 18, 2022
- $\left(4,-\sqrt{6}\right)$
- $\left(7,-2\sqrt{6}\right)$
- $\left(2,3\right)$
- $\left(\sqrt{6},1\right)$
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The Correct Option is A
Solution and Explanation
On solving the equation of line and curve, we get,
$x^{2}-2\left\{\frac{2-2 x}{\sqrt{6}}\right\}^{2}=4$
$\Rightarrow x^{2}-\frac{1}{3} \times 4\left(1+x^{2}-2 x\right)=4$
$\Rightarrow 3 x^{2}-4-4 x^{2}+8 x=12$
$\Rightarrow -x^{2}+8 x-16=0$
$\Rightarrow x^{2}-8 x+16=0$
$\Rightarrow (x-4)^{2}=0 \Rightarrow x=4$
and $\sqrt{6} \cdot y=2-2(4)=-6$
$\Rightarrow y=-\sqrt{6}$
So, point of contact is $(4,-\sqrt{6})$.
$x^{2}-2\left\{\frac{2-2 x}{\sqrt{6}}\right\}^{2}=4$
$\Rightarrow x^{2}-\frac{1}{3} \times 4\left(1+x^{2}-2 x\right)=4$
$\Rightarrow 3 x^{2}-4-4 x^{2}+8 x=12$
$\Rightarrow -x^{2}+8 x-16=0$
$\Rightarrow x^{2}-8 x+16=0$
$\Rightarrow (x-4)^{2}=0 \Rightarrow x=4$
and $\sqrt{6} \cdot y=2-2(4)=-6$
$\Rightarrow y=-\sqrt{6}$
So, point of contact is $(4,-\sqrt{6})$.
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
