1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

Tangents drawn from the point ($$-$$8, 0) to the parabola y2 = 8x touch the parabola at $$P$$ and $$Q.$$ If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :
A
24
B
32
C
48
D
64

Explanation

Equation of the chord of contact PQ is given by : T=0

or T $$ \equiv $$ yy1 $$-$$ 4(x + x1), where (x1, y1) $$ \equiv $$ ($$-$$8, 0)

$$\therefore\,\,\,$$Equation becomes : x = 8

& chord of contact is x = 8

$$\therefore\,\,\,$$ Coordinates of point P and Q are (8, 8) and (8, $$-$$ 8)

and focus of the parabola is F (2, 0)

$$\therefore\,\,\,$$ Area of triangle PQF = $${1 \over 2}$$ $$ \times $$ (8 $$-$$ 2) $$ \times $$ (8 + 8) = 48 sq. units
2
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

A normal to the hyperbola, 4x2 $$-$$ 9y2 = 36 meets the co-ordinate axes $$x$$ and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the ocus of P is :
A
4x2 + 9y2 = 121
B
9x2 + 4y2 = 169
C
4x2 $$-$$ 9y2 = 121
D
9x2 $$-$$ 4y2 = 169

Explanation

Given, 4x2 $$-$$ 9y2 = 36

After differentiating w.r.t.x, we get

4.2x $$-$$ 9.2.y.$${{dy} \over {dx}}$$ = 0

$$ \Rightarrow $$ Slope of tangent = $${{dy} \over {dx}}$$ = $${{4x} \over {9y}}$$

So, slope of normal = $${{ - 9y} \over {4x}}$$

Now, equation of normal at point (x0, y0) is given by

y $$-$$ y0 = $${{ - 9{y_0}} \over {4{x_0}}}$$ (x $$-$$ x0)

As normal intersects X axis at A, Then

A = $$\left( {{{13{x_0}} \over 9},0} \right)$$ and B $$ \equiv $$ $$\left( {0,{{13{y_0}} \over 4}} \right)$$

As OABP is parallelogram

$$ \therefore $$ midpoint of OB $$ \equiv $$ $$\left( {0,{{13{y_0}} \over 8}} \right)$$ $$ \equiv $$ Midpoint of AP

So, P(x, y) $$ \equiv $$ $$\left( {{{ - 13{x_0}} \over 9},{{13{y_0}} \over 4}} \right)$$     ...(i)

$$ \because $$ (x0, y0) lies on hyperbola, therefore

4(x0)2 $$-$$ 9(y0)2 = 36

From equation (i) : x0 = $${{ - 9x} \over {13}}$$ and y0 = $${{4y} \over {13}}$$

From equation (ii), we get

9x2 $$-$$ 4y2 = 169

Hence, locus of point P is : 9x2 $$-$$ 4y2 = 169
3
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

The locus of the point of intersection of the lines, $$\sqrt 2 x - y + 4\sqrt 2 k = 0$$ and $$\sqrt 2 k\,x + k\,y - 4\sqrt 2 = 0$$ (k is any non-zero real parameter), is :
A
an ellipse whose eccentricity is $${1 \over {\sqrt 3 }}.$$
B
an ellipse with length of its major axis $$8\sqrt 2 .$$
C
a hyperbola whose eccentricity is $$\sqrt 3 .$$
D
a hyperbola with length of its transverse axis $$8\sqrt 2 .$$

Explanation

Here, lines are :

$$\sqrt 2 x$$ $$-$$ y + 4$$\sqrt 2 k$$ = 0

$$ \Rightarrow $$$$\,\,\,$$ $$\sqrt 2 x + 4\sqrt 2 k = y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....$$(i)

and   $$\sqrt 2 kx + ky - 4\sqrt 2 = 0\,\,\,\,\,...\left( {ii} \right)$$

Put the value of y from (i) in (ii) we get;

$$ \Rightarrow $$2$$\sqrt 2 $$kx + 4$$\sqrt 2 $$(k2 $$-$$ 1) = 0

$$ \Rightarrow $$ x = $${{2\left( {1 - {k^2}} \right)} \over k}$$, y = $${{2\sqrt 2 \left( {1 + {k^2}} \right)} \over k}$$

$$\therefore\,\,\,$$ $${\left( {{y \over {4\sqrt 2 }}} \right)^2} - {\left( {{x \over 4}} \right)^2} = 1$$

$$\therefore\,\,\,$$ length of transverse axis

2a = 2 $$ \times $$ 4$${\sqrt 2 }$$ = 8$${\sqrt 2 }$$

Hence, the locus is a hyperbola with length of its transverse axis equal to 8$${\sqrt 2 }$$
4
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

Let P be a point on the parabola, x2 = 4y. If the distance of P from the center of the circle, x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is :
A
x + 4y $$-$$ 2 = 0
B
x $$-$$ y + 3 = 0
C
x + y +1 = 0
D
x + 2y = 0

Explanation

Let P(2t, t2) be any point on the parabola.

Center of the given circle C = ($$-$$ g, $$-$$f) = ($$-$$3, 0)

For PC to be minimum, it must be the normal to the parabola at P.

Slope of line PC = $${{{y_2} - {y_1}} \over {{x_2} - {x_1}}}$$ = $${{{t^2} - 0} \over {2t + 3}}$$

Also, slope of tangent to parabola at P = $${{dy} \over {dx}}$$ = $${x \over 2}$$ = t

$$ \therefore $$ Slope of normal = $${{ - 1} \over t}$$

$$ \therefore $$ $${{{t^2} - 0} \over {2t + 3}}$$ = $${{ - 1} \over t}$$

$$ \Rightarrow $$ t3 + 2t + 3 = 0

$$ \Rightarrow $$ (t+1) (t2 $$-$$ t + 3) = 0

$$\therefore\,\,\,$$ Real roots of above equation is

t = $$-$$ 1

Coordinate of P = (2t, t2) = ($$-$$2, 1)

Slope of tangent to parabola at P = t = $$-$$ 1

Therefore, equation of tangent is :

(y $$-$$ 1) = ($$-$$ 1) (x + 2)

$$ \Rightarrow $$ x + y + 1 = 0

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