# Conic Sections - Class 11 - Mathematics

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## Extra Questions - Conic Sections | NCERT | Mathematics | Class 11

If the normal at $\theta$ on the ellipse $5x^{2} + 14y^{2} = 70$ cuts the curve again at a point $2\theta$, then $ \cos \theta = $

A) $\frac{2}{3}$ B) $-\frac{2}{3}$ C) $\frac{1}{3}$ D) $-\frac{1}{3}$

To address the problem, we need to first rewrite the equation of the ellipse in its standard form. Given:

$$ 5x^2 + 14y^2 = 70 $$

we can express it as:

$$ \frac{x^2}{14} + \frac{y^2}{5} = 1 $$

By examining the **equation of the normal** to this ellipse at an angle $\theta$, derived from the parametric equations $x = \sqrt{14}\cos\theta$ and $y = \sqrt{5}\sin\theta$, we have:

$$ \frac{7x}{\cos\theta} - \frac{5y}{\sin\theta} = 0 $$

At the points $\theta$ and $2\theta$, the normal must intersect the ellipse again. If we plug in $x = \sqrt{14}\cos2\theta$ and $y = \sqrt{5}\sin2\theta$ into the equation of the normal, we get:

$$ \frac{7 \sqrt{14} \cos 2\theta}{\cos \theta} - \frac{5 \sqrt{5} \sin 2\theta}{\sin \theta} = 0 $$

Using the **double angle identities**, $\cos2\theta = 2\cos^2\theta - 1$ and $\sin2\theta = 2\sin\theta\cos\theta$, and simplifying further, the equation transforms into:

$$ \sqrt{14} \times 7(2\cos^2\theta - 1) - 5\sqrt{5} \times 2\sin\theta\cos\theta = 0 $$

Simplification leads to a quadratic equation in terms of $\cos\theta$, specifically:

$$ 18\cos^2\theta - 9\cos\theta - 14 = 0 $$

To find $\cos\theta$, solve the quadratic equation using the quadratic formula:

$$ \cos\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

where $a = 18$, $b = -9$, and $c = -14$. Solve this to find:

$$ \cos\theta = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \times 18 \times (-14)}}{2 \times 18} = \frac{9 \pm \sqrt{81 + 1008}}{36} = \frac{9 \pm \sqrt{1089}}{36} $$

The solutions to this are:

$$ \cos\theta = \frac{9 + 33}{36} = \frac{42}{36} = \frac{7}{6} \quad \text{or} \quad \cos\theta = \frac{9 - 33}{36} = \frac{-24}{36} = -\frac{2}{3} $$

Since $\frac{7}{6}$ is not possible for a cosine value (as valid values are between -1 and 1), the only valid answer is:

$$ \cos\theta = -\frac{2}{3} $$

Thus, the correct option is **B) $-\frac{2}{3}$**.

Let two tangents be drawn to the curve $y^{2}-4(x+y)=3 \sin \theta+4 \cos \theta-15$, where $x, y, \theta \in \mathbb{R}$, from the origin with slopes $m_{1}$ and $m_{2}$. If the vertex of the curve is at the maximum distance from the origin, then the value of $\left|\frac{1}{m_{1}m_{2}}\right|$ is:

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$P(\theta)$ and $D\left(\theta+\frac{\pi}{2}\right)$ are two points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Then the locus of the point of intersection of the two tangents at $P$ and $D$ to the ellipse is

(A) $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{1}{4}$

(B) $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=4$

(C) $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=2$

(D) $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{1}{2}$

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The directrix of the parabola $x^{2} - 4x - 8y + 12 = 0$ is

A. $y = 0$ B. $x = 1$ C. $y = -1$ D. $x = -1$

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If (5,12),(24,7) are the foci of the hyperbola passing through the origin, then its eccentricity is:

A. $ \frac{13}{5} $

B. $ \frac{\sqrt{386}}{13} $

C. $\frac{\sqrt{386}}{25}$

D. $\frac{\sqrt{386}}{12}$