A common tangent to the conics $x^{2}=6y$ and $2x^{2}-4y^{2}=9$ is:
- $x-y=\frac{3}{2}$
- $x + y =1$
- $x+y=\frac{9}{2}$
- $x - y = 1$
The Correct Option is A
Solution and Explanation
The correct answer is (A) : \(x-y=\frac{3}{2}\)
Let \(y=(mx+c)\) is tangent to \(x^2=6y\)
Now , \(x^2=6(mx+c)\)
So, \(x^2-6mx-6c=0\)
Put \(D=b^2-4ac=0\)
\(⇒ c=\frac{-3}{2}m^2\)
\(\therefore \) we get \(y=mx-\frac{3}{2}m^2 \) \(.....(1)\)
and given hyperbola equation is \(2x^2-4y^2=9\)
\(⇒\frac{x^2}{\frac{9}{2}}-\frac{y^2}{\frac{9}{4}}=1\) \(....(2)\)
Since, equation (1) is a tangent of equation (2) then \(c^2=a^2m^2-b^2\)
\(⇒\frac{9}{4}m^4=9m^2-\frac{9}{4}\)
\(⇒m^4=2m^2-1\)
\(⇒m^4-2m^2+1=0\)
\(⇒(m^2-1)^2=0\)
\(⇒\)\(m=±1\)
Therefore , for m=1 , equation of tangent is \(x-y=\frac{3}{2}\)
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Conic Sections
When a plane intersects a cone in multiple sections, several types of curves are obtained. These curves can be a circle, an ellipse, a parabola, and a hyperbola. When a plane cuts the cone other than the vertex then the following situations may occur:
Let ‘β’ is the angle made by the plane with the vertical axis of the cone
- When β = 90°, we say the section is a circle
- When α < β < 90°, then the section is an ellipse
- When α = β; then the section is said to as a parabola
- When 0 ≤ β < α; then the section is said to as a hyperbola
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