\(3+4\sqrt 2\)
\(-5+6\sqrt 2\)
\(-4+3\sqrt 2\)
\(7+6\sqrt 2\)
Suppose, tangent to \(y^2 = x\) be \(y=mx+\frac {1}{4m}\)
For tangent to circle,
\(|\frac {\frac 14m}{\sqrt {1+m^2}}|=\sqrt 2\)
\(32m^4 + 32m^2 – 1 = 0\)
According to the Sridharacharya formula,
\(m_2=\frac {−32±\sqrt {(32)^2+4(32)}}{64}\)
\(8m_1m_2=−4+3\sqrt 2\)
So, the correct option is (C): \(−4+3\sqrt 2\)
Let the foci of a hyperbola $ H $ coincide with the foci of the ellipse $ E : \frac{(x - 1)^2}{100} + \frac{(y - 1)^2}{75} = 1 $ and the eccentricity of the hyperbola $ H $ be the reciprocal of the eccentricity of the ellipse $ E $. If the length of the transverse axis of $ H $ is $ \alpha $ and the length of its conjugate axis is $ \beta $, then $ 3\alpha^2 + 2\beta^2 $ is equal to:
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
Read More: Conic Sections