QBManager

Parabola

120230 A chord of the parabola $y = x^{2} - 2 x + 5$ joins the point with the abscissas $x_{1} = 1, x_{2} = 3$ . Then the equation of the tangent to the parabola parallel to the chord is

1

2 x - y + 2 = 0

2

2 x - y + 1 = 0

3

2 x + y + 1 = 0

4

2 x - y + \frac{5}{4} = 0

Explanation:

B Given equation of parabola is
$y = x^{2} - 2 x + 5$
Putting the given value of $x_{1} = 1$ and $x_{2} = 3$ in equation (i), we get, $y_{1} = 4$ and $y_{2} = 8$
$∴$ Points on the parabola are $(1, 4)$ and $(3, 8)$ Equation of the chord of given parabola by joining the points $(1$ , 4 ) and $(3, 8)$ will be
$y - 4 = \frac{8 - 4}{3 - 1} (x - 1)$
$y - 4 = 2 x - 2$
$2 x - y + 2 = 0$
Tangent parallel to this chord will have the slope i.e $\frac{dy}{dx} = 2$
$∴$ Equation of tangent at $(α, β)$ on the curve with slope 2 is $2 x - y + 1 = 0$

UPSEE-2017

Parabola

120231 The point of the contact of the tangent to the parabola $y^{2} = 4 a x$ which makes an angle of $30^{\circ}$ with $X$ -axis, is

1

(3 a, 2 \sqrt{3} a)

2

(2 \sqrt{3} a, 3 a)

3

(\sqrt{3} a, 6 a)

4 None of these

Explanation:

A Given, equation of parabola, $y^{2} = 4 ax$
$m = \tan θ = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$
The equation of tangent at $(h, k)$ to $y^{2} = 4 a x$ is
$yk = 2 a (x + h)$
Comparing; we get, $m = \frac{1}{\sqrt{3}} = \frac{2 a}{k} = k = 2 a \sqrt{3}$
$k = 2 a \sqrt{3} and h = 3 a$

UPSEE-2015

Parabola

120232 A normal is drawn at a point $(x_{1}, y_{1})$ of the parabola $y^{2} = 16 x$ and this normal makes equal angle with both $X$ and $Y$ -axis. Then, point ( $x_{1}$ , $y_{1})$ is

1

(4, - 8)

2

(1, - 4)

3

(4, - 4)

4

(2, - 8)

Explanation:

A Given equation of parabola, $y^{2} = 16 x$
Differentiating both sides, we get
$2 y \frac{d y}{d x} = 16$
$\frac{dy}{dx} = \frac{8}{y}$
$∴$ Slope of at point $(x_{1}, y_{1}), m_{1} = \frac{8}{y_{1}}$
and also slope of normal at point $(x_{1}, y_{1}), m_{2} = \frac{- y_{1}}{8}$
Since, normal makes equal angle with both $X$ and $Y$ axes, then
$m_{2} = \pm 1$
$\frac{- y_{1}}{8} = \pm 1$
$- y_{1} = \pm 8$
When $y_{1} = 8$ , then $x_{1} = 4$
When $y_{1} = - 8$ then $x_{1} = 4$
Hence, the required point is $(4, - 8)$ .

UPSEE-2016

Parabola

120233 The equation of the tangent to the curve $y = 4 e^{- x / 4}$ at the point where the curve crosses $Y$ -axis is equal to

1

3 x + 4 y = 16

2

4 x + y = 4

3

x + y = 4

4

3 x + 4 y = 25

Explanation:

C :
Given, equation of curve $(y) = 4 e^{- x / 4}$
Differentiate equation (i) w.r.t ' $x$ ',
$\frac{d y}{d x} = 4 e^{\frac{- x}{4}} (\frac{- 1}{4})$
$\frac{d y}{d x} = - e^{\frac{- x}{4}}$
According to question, as the curve crosses y-axis,
$x = 0$
$y = 4 e^{- 0}$
$y = 4$
$∴$ Slope of tangents at $(0, 4)$ is,
${[\frac{dy}{dx}]}_{(0, 4)} = m = - e^{- 0}$
$m = - 1$
Equation of tangent of the curve passing through $(0, 4)$ is
$y - 4 = - 1 (x - 0)$
$y - 4 = - x$
$x + y = 4$

BCECE-2016

Parabola

120234 The angle between tangents drawn from the origin to the parabola $y^{2} = 4 a (x - a)$ is

1 0

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{π}{6}

Explanation:

B Given, equation of Parabola,
$y^{2} = 4 a (x - a)$
We know that, $y = mx$
Putting the valve of $y$ from equation (ii) in equation(i),
$m^{2} x^{2} = 4 a (x - a)$
$m^{2} x^{2} - 4 a x + 4 a^{2} = 0$
$D = b^{2} - 4 ac = 0$
$D = (- 4 a)^{2} - 4 (m^{2}) \cdot 4 a^{2}$
$16 a^{2} - 16 a^{2} m^{2} = 0$
$16 a^{2} (1 - m^{2}) = 0$
$1 - m^{2} = 0$
$m^{2} = 1$
$m = \pm 1$
OR
$m_{1} = - 1 or m_{2} = + 1$
$∵ m_{1} \times m_{2} = - 1$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$
$\tan θ = \frac{1 - (- 1)}{1 - 1}$
$\tan θ = \frac{2}{0}$
$\tan θ = \infty$
$\tan θ = \tan 90^{\circ}$
$θ = 90^{\circ} or \frac{π}{2}$

BCECE-2013

Parabola

120230 A chord of the parabola $y = x^{2} - 2 x + 5$ joins the point with the abscissas $x_{1} = 1, x_{2} = 3$ . Then the equation of the tangent to the parabola parallel to the chord is

1

2 x - y + 2 = 0

2

2 x - y + 1 = 0

3

2 x + y + 1 = 0

4

2 x - y + \frac{5}{4} = 0

Explanation:

B Given equation of parabola is
$y = x^{2} - 2 x + 5$
Putting the given value of $x_{1} = 1$ and $x_{2} = 3$ in equation (i), we get, $y_{1} = 4$ and $y_{2} = 8$
$∴$ Points on the parabola are $(1, 4)$ and $(3, 8)$ Equation of the chord of given parabola by joining the points $(1$ , 4 ) and $(3, 8)$ will be
$y - 4 = \frac{8 - 4}{3 - 1} (x - 1)$
$y - 4 = 2 x - 2$
$2 x - y + 2 = 0$
Tangent parallel to this chord will have the slope i.e $\frac{dy}{dx} = 2$
$∴$ Equation of tangent at $(α, β)$ on the curve with slope 2 is $2 x - y + 1 = 0$

UPSEE-2017

Parabola

120231 The point of the contact of the tangent to the parabola $y^{2} = 4 a x$ which makes an angle of $30^{\circ}$ with $X$ -axis, is

1

(3 a, 2 \sqrt{3} a)

2

(2 \sqrt{3} a, 3 a)

3

(\sqrt{3} a, 6 a)

4 None of these

Explanation:

A Given, equation of parabola, $y^{2} = 4 ax$
$m = \tan θ = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$
The equation of tangent at $(h, k)$ to $y^{2} = 4 a x$ is
$yk = 2 a (x + h)$
Comparing; we get, $m = \frac{1}{\sqrt{3}} = \frac{2 a}{k} = k = 2 a \sqrt{3}$
$k = 2 a \sqrt{3} and h = 3 a$

UPSEE-2015

Parabola

120232 A normal is drawn at a point $(x_{1}, y_{1})$ of the parabola $y^{2} = 16 x$ and this normal makes equal angle with both $X$ and $Y$ -axis. Then, point ( $x_{1}$ , $y_{1})$ is

1

(4, - 8)

2

(1, - 4)

3

(4, - 4)

4

(2, - 8)

Explanation:

A Given equation of parabola, $y^{2} = 16 x$
Differentiating both sides, we get
$2 y \frac{d y}{d x} = 16$
$\frac{dy}{dx} = \frac{8}{y}$
$∴$ Slope of at point $(x_{1}, y_{1}), m_{1} = \frac{8}{y_{1}}$
and also slope of normal at point $(x_{1}, y_{1}), m_{2} = \frac{- y_{1}}{8}$
Since, normal makes equal angle with both $X$ and $Y$ axes, then
$m_{2} = \pm 1$
$\frac{- y_{1}}{8} = \pm 1$
$- y_{1} = \pm 8$
When $y_{1} = 8$ , then $x_{1} = 4$
When $y_{1} = - 8$ then $x_{1} = 4$
Hence, the required point is $(4, - 8)$ .

UPSEE-2016

Parabola

120233 The equation of the tangent to the curve $y = 4 e^{- x / 4}$ at the point where the curve crosses $Y$ -axis is equal to

1

3 x + 4 y = 16

2

4 x + y = 4

3

x + y = 4

4

3 x + 4 y = 25

Explanation:

C :
Given, equation of curve $(y) = 4 e^{- x / 4}$
Differentiate equation (i) w.r.t ' $x$ ',
$\frac{d y}{d x} = 4 e^{\frac{- x}{4}} (\frac{- 1}{4})$
$\frac{d y}{d x} = - e^{\frac{- x}{4}}$
According to question, as the curve crosses y-axis,
$x = 0$
$y = 4 e^{- 0}$
$y = 4$
$∴$ Slope of tangents at $(0, 4)$ is,
${[\frac{dy}{dx}]}_{(0, 4)} = m = - e^{- 0}$
$m = - 1$
Equation of tangent of the curve passing through $(0, 4)$ is
$y - 4 = - 1 (x - 0)$
$y - 4 = - x$
$x + y = 4$

BCECE-2016

Parabola

120234 The angle between tangents drawn from the origin to the parabola $y^{2} = 4 a (x - a)$ is

1 0

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{π}{6}

Explanation:

B Given, equation of Parabola,
$y^{2} = 4 a (x - a)$
We know that, $y = mx$
Putting the valve of $y$ from equation (ii) in equation(i),
$m^{2} x^{2} = 4 a (x - a)$
$m^{2} x^{2} - 4 a x + 4 a^{2} = 0$
$D = b^{2} - 4 ac = 0$
$D = (- 4 a)^{2} - 4 (m^{2}) \cdot 4 a^{2}$
$16 a^{2} - 16 a^{2} m^{2} = 0$
$16 a^{2} (1 - m^{2}) = 0$
$1 - m^{2} = 0$
$m^{2} = 1$
$m = \pm 1$
OR
$m_{1} = - 1 or m_{2} = + 1$
$∵ m_{1} \times m_{2} = - 1$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$
$\tan θ = \frac{1 - (- 1)}{1 - 1}$
$\tan θ = \frac{2}{0}$
$\tan θ = \infty$
$\tan θ = \tan 90^{\circ}$
$θ = 90^{\circ} or \frac{π}{2}$

BCECE-2013

Parabola

120230 A chord of the parabola $y = x^{2} - 2 x + 5$ joins the point with the abscissas $x_{1} = 1, x_{2} = 3$ . Then the equation of the tangent to the parabola parallel to the chord is

1

2 x - y + 2 = 0

2

2 x - y + 1 = 0

3

2 x + y + 1 = 0

4

2 x - y + \frac{5}{4} = 0

Explanation:

B Given equation of parabola is
$y = x^{2} - 2 x + 5$
Putting the given value of $x_{1} = 1$ and $x_{2} = 3$ in equation (i), we get, $y_{1} = 4$ and $y_{2} = 8$
$∴$ Points on the parabola are $(1, 4)$ and $(3, 8)$ Equation of the chord of given parabola by joining the points $(1$ , 4 ) and $(3, 8)$ will be
$y - 4 = \frac{8 - 4}{3 - 1} (x - 1)$
$y - 4 = 2 x - 2$
$2 x - y + 2 = 0$
Tangent parallel to this chord will have the slope i.e $\frac{dy}{dx} = 2$
$∴$ Equation of tangent at $(α, β)$ on the curve with slope 2 is $2 x - y + 1 = 0$

UPSEE-2017

Parabola

120231 The point of the contact of the tangent to the parabola $y^{2} = 4 a x$ which makes an angle of $30^{\circ}$ with $X$ -axis, is

1

(3 a, 2 \sqrt{3} a)

2

(2 \sqrt{3} a, 3 a)

3

(\sqrt{3} a, 6 a)

4 None of these

Explanation:

A Given, equation of parabola, $y^{2} = 4 ax$
$m = \tan θ = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$
The equation of tangent at $(h, k)$ to $y^{2} = 4 a x$ is
$yk = 2 a (x + h)$
Comparing; we get, $m = \frac{1}{\sqrt{3}} = \frac{2 a}{k} = k = 2 a \sqrt{3}$
$k = 2 a \sqrt{3} and h = 3 a$

UPSEE-2015

Parabola

120232 A normal is drawn at a point $(x_{1}, y_{1})$ of the parabola $y^{2} = 16 x$ and this normal makes equal angle with both $X$ and $Y$ -axis. Then, point ( $x_{1}$ , $y_{1})$ is

1

(4, - 8)

2

(1, - 4)

3

(4, - 4)

4

(2, - 8)

Explanation:

A Given equation of parabola, $y^{2} = 16 x$
Differentiating both sides, we get
$2 y \frac{d y}{d x} = 16$
$\frac{dy}{dx} = \frac{8}{y}$
$∴$ Slope of at point $(x_{1}, y_{1}), m_{1} = \frac{8}{y_{1}}$
and also slope of normal at point $(x_{1}, y_{1}), m_{2} = \frac{- y_{1}}{8}$
Since, normal makes equal angle with both $X$ and $Y$ axes, then
$m_{2} = \pm 1$
$\frac{- y_{1}}{8} = \pm 1$
$- y_{1} = \pm 8$
When $y_{1} = 8$ , then $x_{1} = 4$
When $y_{1} = - 8$ then $x_{1} = 4$
Hence, the required point is $(4, - 8)$ .

UPSEE-2016

Parabola

120233 The equation of the tangent to the curve $y = 4 e^{- x / 4}$ at the point where the curve crosses $Y$ -axis is equal to

1

3 x + 4 y = 16

2

4 x + y = 4

3

x + y = 4

4

3 x + 4 y = 25

Explanation:

C :
Given, equation of curve $(y) = 4 e^{- x / 4}$
Differentiate equation (i) w.r.t ' $x$ ',
$\frac{d y}{d x} = 4 e^{\frac{- x}{4}} (\frac{- 1}{4})$
$\frac{d y}{d x} = - e^{\frac{- x}{4}}$
According to question, as the curve crosses y-axis,
$x = 0$
$y = 4 e^{- 0}$
$y = 4$
$∴$ Slope of tangents at $(0, 4)$ is,
${[\frac{dy}{dx}]}_{(0, 4)} = m = - e^{- 0}$
$m = - 1$
Equation of tangent of the curve passing through $(0, 4)$ is
$y - 4 = - 1 (x - 0)$
$y - 4 = - x$
$x + y = 4$

BCECE-2016

Parabola

120234 The angle between tangents drawn from the origin to the parabola $y^{2} = 4 a (x - a)$ is

1 0

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{π}{6}

Explanation:

B Given, equation of Parabola,
$y^{2} = 4 a (x - a)$
We know that, $y = mx$
Putting the valve of $y$ from equation (ii) in equation(i),
$m^{2} x^{2} = 4 a (x - a)$
$m^{2} x^{2} - 4 a x + 4 a^{2} = 0$
$D = b^{2} - 4 ac = 0$
$D = (- 4 a)^{2} - 4 (m^{2}) \cdot 4 a^{2}$
$16 a^{2} - 16 a^{2} m^{2} = 0$
$16 a^{2} (1 - m^{2}) = 0$
$1 - m^{2} = 0$
$m^{2} = 1$
$m = \pm 1$
OR
$m_{1} = - 1 or m_{2} = + 1$
$∵ m_{1} \times m_{2} = - 1$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$
$\tan θ = \frac{1 - (- 1)}{1 - 1}$
$\tan θ = \frac{2}{0}$
$\tan θ = \infty$
$\tan θ = \tan 90^{\circ}$
$θ = 90^{\circ} or \frac{π}{2}$

BCECE-2013

Parabola

120230 A chord of the parabola $y = x^{2} - 2 x + 5$ joins the point with the abscissas $x_{1} = 1, x_{2} = 3$ . Then the equation of the tangent to the parabola parallel to the chord is

1

2 x - y + 2 = 0

2

2 x - y + 1 = 0

3

2 x + y + 1 = 0

4

2 x - y + \frac{5}{4} = 0

Explanation:

B Given equation of parabola is
$y = x^{2} - 2 x + 5$
Putting the given value of $x_{1} = 1$ and $x_{2} = 3$ in equation (i), we get, $y_{1} = 4$ and $y_{2} = 8$
$∴$ Points on the parabola are $(1, 4)$ and $(3, 8)$ Equation of the chord of given parabola by joining the points $(1$ , 4 ) and $(3, 8)$ will be
$y - 4 = \frac{8 - 4}{3 - 1} (x - 1)$
$y - 4 = 2 x - 2$
$2 x - y + 2 = 0$
Tangent parallel to this chord will have the slope i.e $\frac{dy}{dx} = 2$
$∴$ Equation of tangent at $(α, β)$ on the curve with slope 2 is $2 x - y + 1 = 0$

UPSEE-2017

Parabola

120231 The point of the contact of the tangent to the parabola $y^{2} = 4 a x$ which makes an angle of $30^{\circ}$ with $X$ -axis, is

1

(3 a, 2 \sqrt{3} a)

2

(2 \sqrt{3} a, 3 a)

3

(\sqrt{3} a, 6 a)

4 None of these

Explanation:

A Given, equation of parabola, $y^{2} = 4 ax$
$m = \tan θ = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$
The equation of tangent at $(h, k)$ to $y^{2} = 4 a x$ is
$yk = 2 a (x + h)$
Comparing; we get, $m = \frac{1}{\sqrt{3}} = \frac{2 a}{k} = k = 2 a \sqrt{3}$
$k = 2 a \sqrt{3} and h = 3 a$

UPSEE-2015

Parabola

120232 A normal is drawn at a point $(x_{1}, y_{1})$ of the parabola $y^{2} = 16 x$ and this normal makes equal angle with both $X$ and $Y$ -axis. Then, point ( $x_{1}$ , $y_{1})$ is

1

(4, - 8)

2

(1, - 4)

3

(4, - 4)

4

(2, - 8)

Explanation:

A Given equation of parabola, $y^{2} = 16 x$
Differentiating both sides, we get
$2 y \frac{d y}{d x} = 16$
$\frac{dy}{dx} = \frac{8}{y}$
$∴$ Slope of at point $(x_{1}, y_{1}), m_{1} = \frac{8}{y_{1}}$
and also slope of normal at point $(x_{1}, y_{1}), m_{2} = \frac{- y_{1}}{8}$
Since, normal makes equal angle with both $X$ and $Y$ axes, then
$m_{2} = \pm 1$
$\frac{- y_{1}}{8} = \pm 1$
$- y_{1} = \pm 8$
When $y_{1} = 8$ , then $x_{1} = 4$
When $y_{1} = - 8$ then $x_{1} = 4$
Hence, the required point is $(4, - 8)$ .

UPSEE-2016

Parabola

120233 The equation of the tangent to the curve $y = 4 e^{- x / 4}$ at the point where the curve crosses $Y$ -axis is equal to

1

3 x + 4 y = 16

2

4 x + y = 4

3

x + y = 4

4

3 x + 4 y = 25

Explanation:

C :
Given, equation of curve $(y) = 4 e^{- x / 4}$
Differentiate equation (i) w.r.t ' $x$ ',
$\frac{d y}{d x} = 4 e^{\frac{- x}{4}} (\frac{- 1}{4})$
$\frac{d y}{d x} = - e^{\frac{- x}{4}}$
According to question, as the curve crosses y-axis,
$x = 0$
$y = 4 e^{- 0}$
$y = 4$
$∴$ Slope of tangents at $(0, 4)$ is,
${[\frac{dy}{dx}]}_{(0, 4)} = m = - e^{- 0}$
$m = - 1$
Equation of tangent of the curve passing through $(0, 4)$ is
$y - 4 = - 1 (x - 0)$
$y - 4 = - x$
$x + y = 4$

BCECE-2016

Parabola

120234 The angle between tangents drawn from the origin to the parabola $y^{2} = 4 a (x - a)$ is

1 0

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{π}{6}

Explanation:

B Given, equation of Parabola,
$y^{2} = 4 a (x - a)$
We know that, $y = mx$
Putting the valve of $y$ from equation (ii) in equation(i),
$m^{2} x^{2} = 4 a (x - a)$
$m^{2} x^{2} - 4 a x + 4 a^{2} = 0$
$D = b^{2} - 4 ac = 0$
$D = (- 4 a)^{2} - 4 (m^{2}) \cdot 4 a^{2}$
$16 a^{2} - 16 a^{2} m^{2} = 0$
$16 a^{2} (1 - m^{2}) = 0$
$1 - m^{2} = 0$
$m^{2} = 1$
$m = \pm 1$
OR
$m_{1} = - 1 or m_{2} = + 1$
$∵ m_{1} \times m_{2} = - 1$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$
$\tan θ = \frac{1 - (- 1)}{1 - 1}$
$\tan θ = \frac{2}{0}$
$\tan θ = \infty$
$\tan θ = \tan 90^{\circ}$
$θ = 90^{\circ} or \frac{π}{2}$

BCECE-2013

Parabola

120230 A chord of the parabola $y = x^{2} - 2 x + 5$ joins the point with the abscissas $x_{1} = 1, x_{2} = 3$ . Then the equation of the tangent to the parabola parallel to the chord is

1

2 x - y + 2 = 0

2

2 x - y + 1 = 0

3

2 x + y + 1 = 0

4

2 x - y + \frac{5}{4} = 0

Explanation:

B Given equation of parabola is
$y = x^{2} - 2 x + 5$
Putting the given value of $x_{1} = 1$ and $x_{2} = 3$ in equation (i), we get, $y_{1} = 4$ and $y_{2} = 8$
$∴$ Points on the parabola are $(1, 4)$ and $(3, 8)$ Equation of the chord of given parabola by joining the points $(1$ , 4 ) and $(3, 8)$ will be
$y - 4 = \frac{8 - 4}{3 - 1} (x - 1)$
$y - 4 = 2 x - 2$
$2 x - y + 2 = 0$
Tangent parallel to this chord will have the slope i.e $\frac{dy}{dx} = 2$
$∴$ Equation of tangent at $(α, β)$ on the curve with slope 2 is $2 x - y + 1 = 0$

UPSEE-2017

Parabola

120231 The point of the contact of the tangent to the parabola $y^{2} = 4 a x$ which makes an angle of $30^{\circ}$ with $X$ -axis, is

1

(3 a, 2 \sqrt{3} a)

2

(2 \sqrt{3} a, 3 a)

3

(\sqrt{3} a, 6 a)

4 None of these

Explanation:

A Given, equation of parabola, $y^{2} = 4 ax$
$m = \tan θ = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$
The equation of tangent at $(h, k)$ to $y^{2} = 4 a x$ is
$yk = 2 a (x + h)$
Comparing; we get, $m = \frac{1}{\sqrt{3}} = \frac{2 a}{k} = k = 2 a \sqrt{3}$
$k = 2 a \sqrt{3} and h = 3 a$

UPSEE-2015

Parabola

120232 A normal is drawn at a point $(x_{1}, y_{1})$ of the parabola $y^{2} = 16 x$ and this normal makes equal angle with both $X$ and $Y$ -axis. Then, point ( $x_{1}$ , $y_{1})$ is

1

(4, - 8)

2

(1, - 4)

3

(4, - 4)

4

(2, - 8)

Explanation:

A Given equation of parabola, $y^{2} = 16 x$
Differentiating both sides, we get
$2 y \frac{d y}{d x} = 16$
$\frac{dy}{dx} = \frac{8}{y}$
$∴$ Slope of at point $(x_{1}, y_{1}), m_{1} = \frac{8}{y_{1}}$
and also slope of normal at point $(x_{1}, y_{1}), m_{2} = \frac{- y_{1}}{8}$
Since, normal makes equal angle with both $X$ and $Y$ axes, then
$m_{2} = \pm 1$
$\frac{- y_{1}}{8} = \pm 1$
$- y_{1} = \pm 8$
When $y_{1} = 8$ , then $x_{1} = 4$
When $y_{1} = - 8$ then $x_{1} = 4$
Hence, the required point is $(4, - 8)$ .

UPSEE-2016

Parabola

120233 The equation of the tangent to the curve $y = 4 e^{- x / 4}$ at the point where the curve crosses $Y$ -axis is equal to

1

3 x + 4 y = 16

2

4 x + y = 4

3

x + y = 4

4

3 x + 4 y = 25

Explanation:

C :
Given, equation of curve $(y) = 4 e^{- x / 4}$
Differentiate equation (i) w.r.t ' $x$ ',
$\frac{d y}{d x} = 4 e^{\frac{- x}{4}} (\frac{- 1}{4})$
$\frac{d y}{d x} = - e^{\frac{- x}{4}}$
According to question, as the curve crosses y-axis,
$x = 0$
$y = 4 e^{- 0}$
$y = 4$
$∴$ Slope of tangents at $(0, 4)$ is,
${[\frac{dy}{dx}]}_{(0, 4)} = m = - e^{- 0}$
$m = - 1$
Equation of tangent of the curve passing through $(0, 4)$ is
$y - 4 = - 1 (x - 0)$
$y - 4 = - x$
$x + y = 4$

BCECE-2016

Parabola

120234 The angle between tangents drawn from the origin to the parabola $y^{2} = 4 a (x - a)$ is

1 0

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{π}{6}

Explanation:

B Given, equation of Parabola,
$y^{2} = 4 a (x - a)$
We know that, $y = mx$
Putting the valve of $y$ from equation (ii) in equation(i),
$m^{2} x^{2} = 4 a (x - a)$
$m^{2} x^{2} - 4 a x + 4 a^{2} = 0$
$D = b^{2} - 4 ac = 0$
$D = (- 4 a)^{2} - 4 (m^{2}) \cdot 4 a^{2}$
$16 a^{2} - 16 a^{2} m^{2} = 0$
$16 a^{2} (1 - m^{2}) = 0$
$1 - m^{2} = 0$
$m^{2} = 1$
$m = \pm 1$
OR
$m_{1} = - 1 or m_{2} = + 1$
$∵ m_{1} \times m_{2} = - 1$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$
$\tan θ = \frac{1 - (- 1)}{1 - 1}$
$\tan θ = \frac{2}{0}$
$\tan θ = \infty$
$\tan θ = \tan 90^{\circ}$
$θ = 90^{\circ} or \frac{π}{2}$

BCECE-2013

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