1

### JEE Main 2016 (Online) 9th April Morning Slot

The minimum distance of a point on the curve y = x2−4 from the origin is :
A
${{\sqrt {19} } \over 2}$
B
$\sqrt {{{15} \over 2}}$
C
${{\sqrt {15} } \over 2}$
D
$\sqrt {{{19} \over 2}}$

## Explanation

Let point on the curve

y = x2 $-$ 4 is ($\alpha$2, $\alpha$2 $-$ 4)

$\therefore$   Distance of the point ($\alpha$2, $\alpha$2 $-$ 4) from origin,

D = $\sqrt {{\alpha ^2} + {{\left( {{\alpha ^2} - 4} \right)}^2}}$

$\Rightarrow$   D2 = $\alpha$2 + $\alpha$4 + 16 $-$ 8$\alpha$2

$=$ $\alpha$4 $-$ 7$\alpha$2 + 16

$\therefore$    ${{d{D^2}} \over {d\alpha }}$ = 4$\alpha$3 $-$ 14$\alpha$

Now, ${{d{D^2}} \over {d\alpha }}$ = 0

$\Rightarrow$   4$\alpha$3 $-$ 14$\alpha$ = 0

$\Rightarrow$   2$\alpha$ (2$\alpha$2 $-$ 7) = 0

$\alpha$ = 0    or   $\alpha$2 = ${7 \over 2}$

${{{d^2}{D^2}} \over {d{\alpha ^2}}} = 12{\alpha ^2} - 14$

$\therefore$   ${\left( {{{{d^2}{D^2}} \over {d{\alpha ^2}}}} \right)_{at\,\,\alpha = 0}} = - 14 < 0$

${\left( {{{{d^2}{D^2}} \over {d{\alpha ^2}}}} \right)_{at\,\,{\alpha ^2} = {7 \over 2}}} = 28 > 0$

$\therefore\,\,\,$ Distance is minimum at $\alpha$2 = ${7 \over 2}$

$\therefore$   Minimum distance

D = $\sqrt {{{49} \over 4} - {{49} \over 4} + 16}$

= ${{\sqrt {15} } \over 2}$
2

### JEE Main 2016 (Online) 10th April Morning Slot

Let C be a curve given by y(x) = 1 + $\sqrt {4x - 3} ,x > {3 \over 4}.$ If P is a point on C, such that the tangent at P has slope ${2 \over 3}$, then a point through which the normal at P passes, is :
A
(2, 3)
B
(4, $-$3)
C
(1, 7)
D
(3, $-$ 4),

## Explanation

Given,

y = 1 + $\sqrt {4x - 3}$

$\therefore$   ${{dy} \over {dx}}$ = ${1 \over {2\sqrt {4x - 3} }} \times 4 = {2 \over 3}$

$\Rightarrow$   4x $-$ 3 = 9

$\Rightarrow$   x = 3

$\therefore$   y = 1 + $\sqrt {12 - 3}$ = 4

$\therefore$   Equation of normal at point P(3,4)

y $-$ 4 = $-$ ${3 \over 2}$ (x $-$ 3)

$\Rightarrow$   2y $-$ 8 = $-$ 3x + 9

$\Rightarrow$    3x + 2y $-$ 17 = 0
3

### JEE Main 2017 (Online) 8th April Morning Slot

The tangent at the point (2, $-$2) to the curve, x2y2 $-$ 2x = 4(1 $-$ y) does not pass through the point :
A
$\left( {4,{1 \over 3}} \right)$
B
(8, 5)
C
($-$4, $-$9)
D
($-$2, $-$7)

## Explanation

As,    ${{dy} \over {dx}}$ = $-$ $\left[ {{{{{\delta f} \over {\delta x}}} \over {{{\delta f} \over {\delta y}}}}} \right]$

${{{\delta f} \over {\delta x}}}$ = y2 $\times$2x $-$ 2

${{{\delta f} \over {\delta y}}}$ = x2 $\times$ 2y + 4

$\therefore\,\,\,$ ${{dy} \over {dx}}$ = $-$ $\left( {{{2x{y^2} - 2} \over {2{x^2}y + 4}}} \right)$

${\left[ {{{dy} \over {dx}}} \right]_{(2, - 2)}}$ = $-$ $\left( {{{2 \times 2 \times 4 - 2} \over {2 \times 4 \times ( - 2) + 4}}} \right)$ = $-$ $\left( {{{14} \over { - 12}}} \right)$ = ${7 \over 6}$

$\therefore\,\,\,$ Slope of tangent to the curve = ${7 \over 6}$

Equation of tangent passes through (2, $-$ 2) is

y + 2 = ${7 \over 6}$ (x $-$ 2)

$\Rightarrow$$\,\,\,$ 7x $-$ 6y = 26 . . . . .(1)

Now put each option in equation (1) and see which one does not satisfy the equation.

By verifying each points you can see ($-$ 2, $-$ 7) does not satisfy the equation.
4

### JEE Main 2017 (Online) 9th April Morning Slot

A tangent to the curve, y = f(x) at P(x, y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 and f(1) = 1, then the curve also passes through the point :
A
$\left( {{1 \over 3},24} \right)$
B
$\left( {{1 \over 2},4} \right)$
C
$\left( {2,{1 \over 8}} \right)$
D
$\left( {3,{1 \over 28}} \right)$