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1

### JEE Advanced 2016 Paper 2 Offline

Let $${F_1}\left( {{x_1},0} \right)$$ and $${F_2}\left( {{x_2},0} \right)$$ for $${{x_1} < 0}$$ and $${{x_2} > 0}$$, be the foci of the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$. Suppose a parabola having vertex at the origin and focus at $${F_2}$$ intersects the ellipse at point $$M$$ in the first quadrant and at point $$N$$ in the fourth quadrant.

The orthocentre of the triangle $${F_1}MN$$ is

A
$$\left( { - {9 \over {10}},0} \right)$$
B
$$\left( { {2 \over {3}},0} \right)$$
C
$$\left( { {9 \over {10}},0} \right)$$
D
$$\left( {{2 \over 3},\sqrt 6 } \right)$$

## Explanation

F1(x, 0) and F2(x2, 0) are the foci of the ellipse:

$${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$

Therefore, a2 = 9 and b2 = 8.

$${b^2} = {a^2}(1 - {e^2})$$

$$1 - {e^2} = {8 \over 9} \Rightarrow {e^2} = 1 - {8 \over 9} = {1 \over 9} \Rightarrow e = {1 \over 3}$$

The focus is

$${F_1}\left( { - 3 \times {1 \over 3},0} \right)$$ and $${F_2}\left( {3 \times {1 \over 3},0} \right)$$

That is, F1($$-$$1, 0) and F2(1, 0). The equation of parabola is

$${y^2} = 4(O{F_2})x$$

$${y^2} = 4x(O{F_2} = 1)$$

The point of intersection of ellipse and parabola is

$${{{x^2}} \over 9} + {{4x} \over 8} = 1 \Rightarrow {{{x^2}} \over 9} + {x \over 2} = 1$$

$$\Rightarrow 2{x^2} + 9x - 18 = 0$$

$$\Rightarrow 2{x^2} + 12x - 3x - 18 = 0$$

$$\Rightarrow 2x(x + 6) - 3(x + 6) = 0$$

$$\Rightarrow x = {3 \over 2}$$ (x $$-$$6 is rejected)

Now, $${y^2}(4){3 \over 2} = 6$$

$$y = \pm \sqrt 6$$

That is, the points M and N are, respectively, $$M\left( {{3 \over 2},\sqrt 6 } \right)$$ and $$N\left( {{3 \over 2}, - \sqrt 6 } \right)$$.

Let the orthocenter be (h, k). The slope of $$OM = {{k - \sqrt 6 } \over {h - (3/2)}}$$

The slope of $$ON = {{\sqrt 6 } \over { - 1 - (3/2)}} = {{ - 2\sqrt 6 } \over 5}$$

Now, $$\left( {{{k - \sqrt 6 } \over {h - (3/2)}}} \right)\left( {{{ - 2\sqrt 6 } \over 5}} \right) = - 1$$

$$2\sqrt 6 k - 12 = 5h - {{15} \over 2}$$

$$5h - 2\sqrt 6 k = {{15} \over 2} - 12 = {{ - 9} \over 2}$$

The slope of $$ON = {{k + \sqrt 6 } \over {h - (3/2)}}$$

The slope of $${F_1}M = {{\sqrt 6 } \over {1 + (3/2)}} = {{2\sqrt 6 } \over 5}$$

$${{k + \sqrt 6 } \over {h - (3/2)}} \times {{2\sqrt 6 } \over 5} = - 1$$

$$2\sqrt 6 k + 12 = - 5h + {{15} \over 2}$$

$$5h + 2\sqrt 6 k = {{15} \over 2} - 12 = {{ - 9} \over 2}$$

$$5h + 2\sqrt 6 k = {{ - 9} \over 2}$$ ....... (1)

$$5h - 2\sqrt 6 k = {{ - 9} \over 2}$$ ........ (2)

Solving Eqs. (1) and (2), we get

$$10h = - 9 \Rightarrow h = {{ - 9} \over {10}}$$ and k = 0

Hence, the orthocentre of the triangle F1MN is $$\left( {{{ - 9} \over {10}},0} \right)$$.

2

### JEE Advanced 2016 Paper 2 Offline

Let $${F_1}\left( {{x_1},0} \right)$$ and $${F_2}\left( {{x_2},0} \right)$$ for $${{x_1} < 0}$$ and $${{x_2} > 0}$$, be the foci of the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$. Suppose a parabola having vertex at the origin and focus at $${F_2}$$ intersects the ellipse at point $$M$$ in the first quadrant and at point $$N$$ in the fourth quadrant.

If the tangents to the ellipse at $$M$$ and $$N$$ meet at $$R$$ and the normal to the parabola at $$M$$ meets the $$x$$-axis at $$Q$$, then the ratio of area of the triangle $$MQR$$ to area of the quadrilateral $$M{F_1}N{F_2}$$is

A
$$3:4$$
B
$$4:5$$
C
$$5:8$$
D
$$2:3$$

## Explanation

Equation of tangent at M(3/2, $$\sqrt6$$) to $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$ is

$${3 \over 2}.{x \over 9} + \sqrt 6 .{y \over 8} = 1$$ ....... (i)

which intersect X-axis at (6, 0).

Also, equation of tangent at N(3/2, $$-$$$$\sqrt6$$) is

$${3 \over 2}.{x \over 9} - \sqrt 6 .{y \over 8} = 1$$ ....... (ii)

Eqs. (i) and (ii) intersect on X-axis at R(6, 0). ........ (iii)

Also, normal at $$M(3/2,\sqrt 6 )$$ is

$$y - \sqrt 6 = {{ - \sqrt 6 } \over 2}\left( {x - {3 \over 2}} \right)$$

On solving with y = 0, we get Q(7/2, 0) ....... (iv) The area of MQR is

$$\left| {{1 \over 2}} \right|\left| {\matrix{ {3/2} & {\sqrt 6 } & 1 \cr 6 & 0 & 1 \cr {7/2} & 0 & 1 \cr } } \right| = \left| {{{\sqrt 6 } \over 2}\left( {6 - {7 \over 2}} \right)} \right| = {{5\sqrt 6 } \over 4}$$

The area of the quadrilateral MF1NF2 is

$$2(\Delta {m_1}{F_1}{F_2}) = 2\sqrt 6$$

and the required ratio is

$${{5\sqrt 6 } \over {4\,.\,2\sqrt 6 }} = {5 \over 8}$$

3

### JEE Advanced 2014 Paper 2 Offline

Let $$a, r, s, t$$ be nonzero real numbers. Let $$P\,\,\left( {a{t^2},2at} \right),\,\,Q,\,\,\,R\,\,\left( {a{r^2},2ar} \right)$$ and $$S\,\,\left( {a{s^2},2as} \right)$$ be distinct points on the parabola $${y^2} = 4ax$$. Suppose that $$PQ$$ is the focal chord and lines $$QR$$ and $$PK$$ are parallel, where $$K$$ is the point $$(2a,0)$$

If $$st=1$$, then the tangent at $$P$$ and the normal at $$S$$ to the parabola meet at a point whose ordinate is

A
$${{{{\left( {{t^2} + 1} \right)}^2}} \over {2{t^3}}}$$
B
$${{a{{\left( {{t^2} + 1} \right)}^2}} \over {2{t^3}}}$$
C
$${{a{{\left( {{t^2} + 1} \right)}^2}} \over {{t^3}}}$$
D
$${{a{{\left( {{t^2} + 2} \right)}^2}} \over {{t^3}}}$$
4

### JEE Advanced 2014 Paper 2 Offline

Let $$a, r, s, t$$ be nonzero real numbers. Let $$P\,\,\left( {a{t^2},2at} \right),\,\,Q,\,\,\,R\,\,\left( {a{r^2},2ar} \right)$$ and $$S\,\,\left( {a{s^2},2as} \right)$$ be distinct points on the parabola $${y^2} = 4ax$$. Suppose that $$PQ$$ is the focal chord and lines $$QR$$ and $$PK$$ are parallel, where $$K$$ is the point $$(2a,0)$$

The value of $$r$$ is

A
$$- {1 \over t}$$
B
$${{{t^2} + 1} \over t}$$
C
$${1 \over t}$$
D
$${{{t^2} - 1} \over t}$$

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