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### IIT-JEE 2012 Paper 1 Offline

Numerical
Let $$S$$ be the focus of the parabola $${y^2} = 8x$$ and let $$PQ$$ be the common chord of the circle $${x^2} + {y^2} - 2x - 4y = 0$$ and the given parabola. The area of the triangle $$PQS$$ is

2

### IIT-JEE 2011 Paper 1 Offline

Numerical
Consider the parabola $${y^2} = 8x$$. Let $${\Delta _1}$$ be the area of the triangle formed by the end points of its latus rectum and the point $$P\left( {{1 \over 2},2} \right)$$ on the parabola and $${\Delta _2}$$ be the area of the triangle formed by drawing tangents at $$P$$ and at the end points of the latus rectum. Then $${{{\Delta _1}} \over {{\Delta _2}}}$$ is

3

### IIT-JEE 2010 Paper 1 Offline

Numerical

The line $$2x + y = 1$$ is tangent to the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$.

If this line passes through the point of intersection of the nearest directrix and the $$x$$-axis, then the eccentricity of the hyperbola is

## Explanation

On substituting $$\left( {{a \over e},0} \right)$$ in $$y = - 2x + 1$$,

we get

$$0 = - {{2a} \over e} + 1$$

$$\Rightarrow {a \over e} = {1 \over 2}$$

Also, $$y = - 2x + 1$$ is tangent to hyperbola

$$\therefore$$ $$1 = 4{a^2} - {b^2}$$

$$\Rightarrow {1 \over {{a^2}}} = 4 - ({e^2} - 1)$$

$$\Rightarrow {4 \over {{e^2}}} = 5 - {e^2}$$

$$\Rightarrow {e^4} - 5{e^2} + 4 = 0$$

$$\Rightarrow ({e^2} - 4)({e^2} - 1) = 0$$

$$\Rightarrow$$ e = 2, e = 1

e = 1 gives the conic as parabola. But conic is given as hyperbola, hence e = 2.

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