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Parabola

120992 The equation of the tangent to the parabola $y^{2}$ $= 4 x$ inclined at an angle of $\frac{π}{4}$ to the positive direction of $x$ -axis is

1

x + y - 4 = 0

2

x - y + 4 = 0

3

x - y - 1 = 0

4

x - y + 1 = 0

Explanation:

D Given parabola equation, $y^{2} = 4 x$
Differentiate both sides w.r.t ' $x$ ',
$2 y \frac{d y}{d x} = 4$
$\frac{d y}{d x} = \frac{2}{y}$
Since, the tangent is inclined at $\frac{π}{4}$ in positive direction of $x$ -axis, $\frac{2}{y^{'}} = \tan (\frac{π}{4}) = 1$
$y^{'} = 2$
Correspondingly $x^{'} = 1$
Thus, we need a line having slope 1 and passing through $(1, 2)$
which is $y - x - 1 = 0$ or $x - y + 1 = 0$ .

Karnataka CET-2013

Parabola

120993 The locus of the point of intersection of the tangents drawn at the ends of a focal chord of the parabola $x^{2} = - 8 y$ is

1

x = - 2

2

x = 2

3

y = - 2

4

y = 2

Explanation:

D Given, parabola $x^{2} = - 8 y$
Comparing, $x^{2} = 4$ ay
$4 a = - 8$
$a = - 2$
Let the parametric coordinate of parabola
$x^{2} = - 8 y is, P \to (4 t, - 2 t^{2})$
and the other coordinate of latusrectum is
original image
$R \Rightarrow (\frac{- 4}{t}, \frac{- 2}{t^{2}})$
Now, the equation of tangent of parabola $x^{2} = - 8 y$
X. $x_{1} = - 4 (y + y_{1})$
At $P$
$x \cdot x_{1} = - 4 (y + y_{1})$
$x (4 t) = - 4 (y - 2 t^{2})$
$x t = - y + 2 t^{2}$
$x t + y = 2 t^{2}$
$x (\frac{- 4}{t}) = - 4 (y - \frac{2}{t^{2}})$
$\frac{x}{t} = y - \frac{2}{t^{2}}$
$x t = y t^{2} - 2$
$x t - y t^{2} = - 2$
On solving equation (i) and (iii)
$x t^{3} + y t^{2} = 2 t^{4}$
$x t - y t^{2} = - 2$
$x t (1 + t^{2}) = - 2 (1 - t^{4})$
$t x = - 2 (1 - t^{2})$
From equation (ii)
$- 2 (1 - t^{2}) + y = 2 t^{2}$
$- 2 + 2 t^{2} + y = 2 t^{2}$
$y = 2$ Hence, the intersection point of both tangent lying on $Q$ . Hence, $y = 2$ which is the required locus.

Karnataka CET-2010

Parabola

120994 The locus of the point of intersection of two tangents to the parabola $y^{2} = 4 ax$ , which are at right angle to one another is

1

x^{2} + y^{2} = a^{2}

2

a y^{2} = x

3

x + a = 0

4

x + y \pm a = 0

Explanation:

C Let the two tangent to the parabola $y^{2} = 4 ax$ be PT and QT which are at right angle to one another at $T (e, f)$ . Then we have to find the locus of $T (e, f)$ .
We know that $y = m x + \frac{a}{m}$ , where $m$ is the slope is the equation of tangent to the parabola $y^{2} = 4 ax$
original image
Since, this tangent to the parabola will pass through $T (e, f)$ so
$f = me + \frac{a}{m}; or m^{2} e - mf + a = 0$
This is a quadratic equation in $m$ . So we have two roots, say $m_{1}$ and $m_{2}$ , then
$m_{1} + m_{2} = \frac{f}{e}, and m_{1} \times m_{2} = \frac{a}{e}$
$∵$ The two tangents intersect at right angle so
$m_{1} \cdot m_{2} = - 1 or \frac{a}{e} = - 1 or h + a = 0$ The locus of $T (e, f)$ is $x + a = 0$ , which is the equation of directrix.

BITSAT-2016

Parabola

120995 The equation of one of the common tangents to the parabola $y^{2} = 8 x$ and $x^{2} + y^{2} - 12 x + 4 = 0$ is

1

y = - x + 2

2

y = x - 2

3

y = x + 2

4 None of these

Explanation:

C
Given,
Tangent to parabola $y^{2} = 8 x is$
$y = mx + \frac{2}{m}$
It touches the circle $x^{2} + y^{2} - 12 x + 4 = 0$ , if the length perpendicular from the centre $(6, 0)$ is equal to radius.
$\pm \sqrt{32} unit.$
$\frac{6 m + \frac{2}{m}}{\sqrt{m^{2} + 1}} = \pm \sqrt{32} \Rightarrow {(3 m + \frac{1}{m})}^{2} = 8 (m^{2} + 1)$
${(3 m^{2} + 1)}^{2} = 8 (m^{4} + m^{2})$
$m^{4} - 2 m^{2} + 1 = 0 \Rightarrow m \pm 1$
Hence, the required tangents are $y = x + 2$ and $y = - x - 2 .$

BITSAT-2017

Parabola

120992 The equation of the tangent to the parabola $y^{2}$ $= 4 x$ inclined at an angle of $\frac{π}{4}$ to the positive direction of $x$ -axis is

1

x + y - 4 = 0

2

x - y + 4 = 0

3

x - y - 1 = 0

4

x - y + 1 = 0

Explanation:

D Given parabola equation, $y^{2} = 4 x$
Differentiate both sides w.r.t ' $x$ ',
$2 y \frac{d y}{d x} = 4$
$\frac{d y}{d x} = \frac{2}{y}$
Since, the tangent is inclined at $\frac{π}{4}$ in positive direction of $x$ -axis, $\frac{2}{y^{'}} = \tan (\frac{π}{4}) = 1$
$y^{'} = 2$
Correspondingly $x^{'} = 1$
Thus, we need a line having slope 1 and passing through $(1, 2)$
which is $y - x - 1 = 0$ or $x - y + 1 = 0$ .

Karnataka CET-2013

Parabola

120993 The locus of the point of intersection of the tangents drawn at the ends of a focal chord of the parabola $x^{2} = - 8 y$ is

1

x = - 2

2

x = 2

3

y = - 2

4

y = 2

Explanation:

D Given, parabola $x^{2} = - 8 y$
Comparing, $x^{2} = 4$ ay
$4 a = - 8$
$a = - 2$
Let the parametric coordinate of parabola
$x^{2} = - 8 y is, P \to (4 t, - 2 t^{2})$
and the other coordinate of latusrectum is
original image
$R \Rightarrow (\frac{- 4}{t}, \frac{- 2}{t^{2}})$
Now, the equation of tangent of parabola $x^{2} = - 8 y$
X. $x_{1} = - 4 (y + y_{1})$
At $P$
$x \cdot x_{1} = - 4 (y + y_{1})$
$x (4 t) = - 4 (y - 2 t^{2})$
$x t = - y + 2 t^{2}$
$x t + y = 2 t^{2}$
$x (\frac{- 4}{t}) = - 4 (y - \frac{2}{t^{2}})$
$\frac{x}{t} = y - \frac{2}{t^{2}}$
$x t = y t^{2} - 2$
$x t - y t^{2} = - 2$
On solving equation (i) and (iii)
$x t^{3} + y t^{2} = 2 t^{4}$
$x t - y t^{2} = - 2$
$x t (1 + t^{2}) = - 2 (1 - t^{4})$
$t x = - 2 (1 - t^{2})$
From equation (ii)
$- 2 (1 - t^{2}) + y = 2 t^{2}$
$- 2 + 2 t^{2} + y = 2 t^{2}$
$y = 2$ Hence, the intersection point of both tangent lying on $Q$ . Hence, $y = 2$ which is the required locus.

Karnataka CET-2010

Parabola

120994 The locus of the point of intersection of two tangents to the parabola $y^{2} = 4 ax$ , which are at right angle to one another is

1

x^{2} + y^{2} = a^{2}

2

a y^{2} = x

3

x + a = 0

4

x + y \pm a = 0

Explanation:

C Let the two tangent to the parabola $y^{2} = 4 ax$ be PT and QT which are at right angle to one another at $T (e, f)$ . Then we have to find the locus of $T (e, f)$ .
We know that $y = m x + \frac{a}{m}$ , where $m$ is the slope is the equation of tangent to the parabola $y^{2} = 4 ax$
original image
Since, this tangent to the parabola will pass through $T (e, f)$ so
$f = me + \frac{a}{m}; or m^{2} e - mf + a = 0$
This is a quadratic equation in $m$ . So we have two roots, say $m_{1}$ and $m_{2}$ , then
$m_{1} + m_{2} = \frac{f}{e}, and m_{1} \times m_{2} = \frac{a}{e}$
$∵$ The two tangents intersect at right angle so
$m_{1} \cdot m_{2} = - 1 or \frac{a}{e} = - 1 or h + a = 0$ The locus of $T (e, f)$ is $x + a = 0$ , which is the equation of directrix.

BITSAT-2016

Parabola

120995 The equation of one of the common tangents to the parabola $y^{2} = 8 x$ and $x^{2} + y^{2} - 12 x + 4 = 0$ is

1

y = - x + 2

2

y = x - 2

3

y = x + 2

4 None of these

Explanation:

C
Given,
Tangent to parabola $y^{2} = 8 x is$
$y = mx + \frac{2}{m}$
It touches the circle $x^{2} + y^{2} - 12 x + 4 = 0$ , if the length perpendicular from the centre $(6, 0)$ is equal to radius.
$\pm \sqrt{32} unit.$
$\frac{6 m + \frac{2}{m}}{\sqrt{m^{2} + 1}} = \pm \sqrt{32} \Rightarrow {(3 m + \frac{1}{m})}^{2} = 8 (m^{2} + 1)$
${(3 m^{2} + 1)}^{2} = 8 (m^{4} + m^{2})$
$m^{4} - 2 m^{2} + 1 = 0 \Rightarrow m \pm 1$
Hence, the required tangents are $y = x + 2$ and $y = - x - 2 .$

BITSAT-2017

Parabola

120992 The equation of the tangent to the parabola $y^{2}$ $= 4 x$ inclined at an angle of $\frac{π}{4}$ to the positive direction of $x$ -axis is

1

x + y - 4 = 0

2

x - y + 4 = 0

3

x - y - 1 = 0

4

x - y + 1 = 0

Explanation:

D Given parabola equation, $y^{2} = 4 x$
Differentiate both sides w.r.t ' $x$ ',
$2 y \frac{d y}{d x} = 4$
$\frac{d y}{d x} = \frac{2}{y}$
Since, the tangent is inclined at $\frac{π}{4}$ in positive direction of $x$ -axis, $\frac{2}{y^{'}} = \tan (\frac{π}{4}) = 1$
$y^{'} = 2$
Correspondingly $x^{'} = 1$
Thus, we need a line having slope 1 and passing through $(1, 2)$
which is $y - x - 1 = 0$ or $x - y + 1 = 0$ .

Karnataka CET-2013

Parabola

120993 The locus of the point of intersection of the tangents drawn at the ends of a focal chord of the parabola $x^{2} = - 8 y$ is

1

x = - 2

2

x = 2

3

y = - 2

4

y = 2

Explanation:

D Given, parabola $x^{2} = - 8 y$
Comparing, $x^{2} = 4$ ay
$4 a = - 8$
$a = - 2$
Let the parametric coordinate of parabola
$x^{2} = - 8 y is, P \to (4 t, - 2 t^{2})$
and the other coordinate of latusrectum is
original image
$R \Rightarrow (\frac{- 4}{t}, \frac{- 2}{t^{2}})$
Now, the equation of tangent of parabola $x^{2} = - 8 y$
X. $x_{1} = - 4 (y + y_{1})$
At $P$
$x \cdot x_{1} = - 4 (y + y_{1})$
$x (4 t) = - 4 (y - 2 t^{2})$
$x t = - y + 2 t^{2}$
$x t + y = 2 t^{2}$
$x (\frac{- 4}{t}) = - 4 (y - \frac{2}{t^{2}})$
$\frac{x}{t} = y - \frac{2}{t^{2}}$
$x t = y t^{2} - 2$
$x t - y t^{2} = - 2$
On solving equation (i) and (iii)
$x t^{3} + y t^{2} = 2 t^{4}$
$x t - y t^{2} = - 2$
$x t (1 + t^{2}) = - 2 (1 - t^{4})$
$t x = - 2 (1 - t^{2})$
From equation (ii)
$- 2 (1 - t^{2}) + y = 2 t^{2}$
$- 2 + 2 t^{2} + y = 2 t^{2}$
$y = 2$ Hence, the intersection point of both tangent lying on $Q$ . Hence, $y = 2$ which is the required locus.

Karnataka CET-2010

Parabola

120994 The locus of the point of intersection of two tangents to the parabola $y^{2} = 4 ax$ , which are at right angle to one another is

1

x^{2} + y^{2} = a^{2}

2

a y^{2} = x

3

x + a = 0

4

x + y \pm a = 0

Explanation:

C Let the two tangent to the parabola $y^{2} = 4 ax$ be PT and QT which are at right angle to one another at $T (e, f)$ . Then we have to find the locus of $T (e, f)$ .
We know that $y = m x + \frac{a}{m}$ , where $m$ is the slope is the equation of tangent to the parabola $y^{2} = 4 ax$
original image
Since, this tangent to the parabola will pass through $T (e, f)$ so
$f = me + \frac{a}{m}; or m^{2} e - mf + a = 0$
This is a quadratic equation in $m$ . So we have two roots, say $m_{1}$ and $m_{2}$ , then
$m_{1} + m_{2} = \frac{f}{e}, and m_{1} \times m_{2} = \frac{a}{e}$
$∵$ The two tangents intersect at right angle so
$m_{1} \cdot m_{2} = - 1 or \frac{a}{e} = - 1 or h + a = 0$ The locus of $T (e, f)$ is $x + a = 0$ , which is the equation of directrix.

BITSAT-2016

Parabola

120995 The equation of one of the common tangents to the parabola $y^{2} = 8 x$ and $x^{2} + y^{2} - 12 x + 4 = 0$ is

1

y = - x + 2

2

y = x - 2

3

y = x + 2

4 None of these

Explanation:

C
Given,
Tangent to parabola $y^{2} = 8 x is$
$y = mx + \frac{2}{m}$
It touches the circle $x^{2} + y^{2} - 12 x + 4 = 0$ , if the length perpendicular from the centre $(6, 0)$ is equal to radius.
$\pm \sqrt{32} unit.$
$\frac{6 m + \frac{2}{m}}{\sqrt{m^{2} + 1}} = \pm \sqrt{32} \Rightarrow {(3 m + \frac{1}{m})}^{2} = 8 (m^{2} + 1)$
${(3 m^{2} + 1)}^{2} = 8 (m^{4} + m^{2})$
$m^{4} - 2 m^{2} + 1 = 0 \Rightarrow m \pm 1$
Hence, the required tangents are $y = x + 2$ and $y = - x - 2 .$

BITSAT-2017

Parabola

120992 The equation of the tangent to the parabola $y^{2}$ $= 4 x$ inclined at an angle of $\frac{π}{4}$ to the positive direction of $x$ -axis is

1

x + y - 4 = 0

2

x - y + 4 = 0

3

x - y - 1 = 0

4

x - y + 1 = 0

Explanation:

D Given parabola equation, $y^{2} = 4 x$
Differentiate both sides w.r.t ' $x$ ',
$2 y \frac{d y}{d x} = 4$
$\frac{d y}{d x} = \frac{2}{y}$
Since, the tangent is inclined at $\frac{π}{4}$ in positive direction of $x$ -axis, $\frac{2}{y^{'}} = \tan (\frac{π}{4}) = 1$
$y^{'} = 2$
Correspondingly $x^{'} = 1$
Thus, we need a line having slope 1 and passing through $(1, 2)$
which is $y - x - 1 = 0$ or $x - y + 1 = 0$ .

Karnataka CET-2013

Parabola

120993 The locus of the point of intersection of the tangents drawn at the ends of a focal chord of the parabola $x^{2} = - 8 y$ is

1

x = - 2

2

x = 2

3

y = - 2

4

y = 2

Explanation:

D Given, parabola $x^{2} = - 8 y$
Comparing, $x^{2} = 4$ ay
$4 a = - 8$
$a = - 2$
Let the parametric coordinate of parabola
$x^{2} = - 8 y is, P \to (4 t, - 2 t^{2})$
and the other coordinate of latusrectum is
original image
$R \Rightarrow (\frac{- 4}{t}, \frac{- 2}{t^{2}})$
Now, the equation of tangent of parabola $x^{2} = - 8 y$
X. $x_{1} = - 4 (y + y_{1})$
At $P$
$x \cdot x_{1} = - 4 (y + y_{1})$
$x (4 t) = - 4 (y - 2 t^{2})$
$x t = - y + 2 t^{2}$
$x t + y = 2 t^{2}$
$x (\frac{- 4}{t}) = - 4 (y - \frac{2}{t^{2}})$
$\frac{x}{t} = y - \frac{2}{t^{2}}$
$x t = y t^{2} - 2$
$x t - y t^{2} = - 2$
On solving equation (i) and (iii)
$x t^{3} + y t^{2} = 2 t^{4}$
$x t - y t^{2} = - 2$
$x t (1 + t^{2}) = - 2 (1 - t^{4})$
$t x = - 2 (1 - t^{2})$
From equation (ii)
$- 2 (1 - t^{2}) + y = 2 t^{2}$
$- 2 + 2 t^{2} + y = 2 t^{2}$
$y = 2$ Hence, the intersection point of both tangent lying on $Q$ . Hence, $y = 2$ which is the required locus.

Karnataka CET-2010

Parabola

120994 The locus of the point of intersection of two tangents to the parabola $y^{2} = 4 ax$ , which are at right angle to one another is

1

x^{2} + y^{2} = a^{2}

2

a y^{2} = x

3

x + a = 0

4

x + y \pm a = 0

Explanation:

C Let the two tangent to the parabola $y^{2} = 4 ax$ be PT and QT which are at right angle to one another at $T (e, f)$ . Then we have to find the locus of $T (e, f)$ .
We know that $y = m x + \frac{a}{m}$ , where $m$ is the slope is the equation of tangent to the parabola $y^{2} = 4 ax$
original image
Since, this tangent to the parabola will pass through $T (e, f)$ so
$f = me + \frac{a}{m}; or m^{2} e - mf + a = 0$
This is a quadratic equation in $m$ . So we have two roots, say $m_{1}$ and $m_{2}$ , then
$m_{1} + m_{2} = \frac{f}{e}, and m_{1} \times m_{2} = \frac{a}{e}$
$∵$ The two tangents intersect at right angle so
$m_{1} \cdot m_{2} = - 1 or \frac{a}{e} = - 1 or h + a = 0$ The locus of $T (e, f)$ is $x + a = 0$ , which is the equation of directrix.

BITSAT-2016

Parabola

120995 The equation of one of the common tangents to the parabola $y^{2} = 8 x$ and $x^{2} + y^{2} - 12 x + 4 = 0$ is

1

y = - x + 2

2

y = x - 2

3

y = x + 2

4 None of these

Explanation:

C
Given,
Tangent to parabola $y^{2} = 8 x is$
$y = mx + \frac{2}{m}$
It touches the circle $x^{2} + y^{2} - 12 x + 4 = 0$ , if the length perpendicular from the centre $(6, 0)$ is equal to radius.
$\pm \sqrt{32} unit.$
$\frac{6 m + \frac{2}{m}}{\sqrt{m^{2} + 1}} = \pm \sqrt{32} \Rightarrow {(3 m + \frac{1}{m})}^{2} = 8 (m^{2} + 1)$
${(3 m^{2} + 1)}^{2} = 8 (m^{4} + m^{2})$
$m^{4} - 2 m^{2} + 1 = 0 \Rightarrow m \pm 1$
Hence, the required tangents are $y = x + 2$ and $y = - x - 2 .$

BITSAT-2017