List of top Mathematics Questions on sections of a cone asked in KEAM

Let $R(-2,-2)$ be a point and let $\dfrac{(x-3)^2}{25} + \dfrac{(y+2)^2}{16} = 1$ be an ellipse. If $S$ and $T$ are the foci of the ellipse, then $RS + RT$ is equal to:

The equation of the latus rectum of the parabola $y^2 + 8x + 4y + 12 = 0$ is

The vertex of a parabola is at $(2,-5)$ and the focus is at $(5,-5)$. The equation of the parabola is

Let $P = \left(\frac{15}{2}(\csc \theta + \sin \theta), \; 8(\csc \theta - \sin \theta)\right)$, where $\theta$ is a variable parameter. Then the locus of $P$ is

The distance between the foci of the ellipse \(\frac{(x + 2)^{2{9} + \frac{(y - 1)^{2}}{4} = 1\) is}

The axis of the parabola \(x^{2} + 6x + 4y + 5 = 0\) is

The distance between the foci of the hyperbola \( x^2-4y^2=16 \), is

If the length of the latus rectum of an ellipse is one-fourth of the major axis, then the eccentricity of the ellipse is

If one end of the latus rectum of the parabola \(y^2=16x\) is \((4,8)\), then the coordinates of the other end of the latus rectum are

The eccentricity of the hyperbola $\frac{(x-1)^{2}}{25}-\frac{(y+2)^{2}}{11}=1$ is:

Let $P$ be any point on the ellipse $4(x+2)^{2}+9(y-4)^{2}=144$. If $F_{1}$ and $F_{2}$ are the foci of the ellipse, then $F_{1}P+F_{2}P=$

Let $y^{2}=8x$ be the equation of a parabola. Which one of the following is an arbitrary point on the parabola?

The eccentricity of the hyperbola \( \frac{(2x-6)^2}{2} - \frac{(4y+7)^2}{16} = 1 \) is

The vertex of the parabola \( 4y = x^2 - 6x + 17 \) is

The length of major axis and minor axis of an ellipse are, respectively, \( m \) and \( n \). If \( m^2 - n^2 = 45 \) and the eccentricity of the ellipse is \( \frac{\sqrt{5}}{3} \), then the length of the major axis is

The focus of the parabola \( x^2 - 4x + 8y + 4 = 0 \) is

The coordinates of the focus and the vertex of a parabola, respectively, are $(-1,\,4)$ and $(3,\,4)$. Then the equation of the parabola is

The line segment joining the points $(-3,\,1)$ and $(1,\,1)$ is the transverse axis of a hyperbola. If the length of the conjugate axis is 4, then the equation of the hyperbola is

The length of the latus rectum of the parabola \( (x+2)^2 = -14(y-5) \) is

One of the foci of the hyperbola \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \) is

The equation of the directrix of the parabola $y^2 + 4y + 4x + 2 = 0$ is

The foci of the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$ are

The axis of the parabola \( x^2 + 6x + 4y + 5 = 0 \) is: