QBManager

Application of Derivatives

85418 What is the $x$ -coordinate of the point on the curve $f (x) = \sqrt{x} (7 x - 6)$ , where the tangent is parallel to $x$ -axis

1

- \frac{1}{3}

2

\frac{2}{7}

3

\frac{6}{7}

4

\frac{1}{2}

Explanation:

(B) :Given,
$f (x) = \sqrt{x} (7 x - 6)$
$f (x) = 7 x^{3 / 2} - 6 x^{1 / 2}$
On differentiation both sides w.r.t.x, we get-
$f^{'} (x) = 7 \times \frac{3}{2} x^{1 / 2} - 6 \times \frac{1}{2} x^{- 1 / 2}$
When tangent is parallel to $x$ axis $f^{'} (x) = 0$
$\frac{21}{2} x^{1 / 2} - 3 x^{- 1 / 2} = 0 \Rightarrow \frac{21}{2} \sqrt{x} = \frac{3}{\sqrt{x}}$
$7 x = 2 \Rightarrow x = \frac{2}{7}$

BITSAT-2016

Application of Derivatives

85419 For the curve $x^{2} + 4 x y + 8 y^{2} = 64$ the tangents are parallel to the $x$ -axis only at the points

1

(0, 2 \sqrt{2})

and

(0, - 2 \sqrt{2})

2

(8, - 4)

and

(- 8, 4)

3

(8 \sqrt{2}, - 2 \sqrt{2})

and

(- 8 \sqrt{2}, 2 \sqrt{2})

4

(9, 0)

and

(- 8, 0)

Explanation:

(B) : Given,
$x^{2} + 4 x y + 8 y^{2} = 64$
On differentiating both sides w.r.t to $x$ we get -
$2 x + 4 (y + x \frac{d y}{d x}) + 16 y \frac{d y}{d x} = 0$
$2 x + 4 y + (4 x + 16 y) \frac{d y}{d x} = 0$
$\frac{d y}{d x} = - \frac{(x + 2 y)}{2 (x + 4 y)}$
Since, tangent are parallel to $x$ -axis only.
$\begin{aligned} \frac{d y}{d x} = 0 \Rightarrow 0 = - \frac{(x + 2 y)}{2 (x + 4 y)} \\ x + 2 y = 0 \end{aligned}$
On putting the values of $x$ from equation (i) in (ii), we get $4 y^{2} - 8 y^{2} + 8 y^{2} = 64$
$4 y^{2} = 64$
$y^{2} = 16$
$y = \pm 4$
On putting the value $y$ in equation (ii)
$x + 2 y = 0$
$x + 2 (4) = 0$
$x = - 8, y = 4$
$x = + 8, y = - 4$
Hence, required point are $(- 8, 4)$ and $(8, - 4)$

WB JEE-2013

Application of Derivatives

85420 If the equation of one tangent to the circle with centre at $(2, - 1)$ from the origin is $3 x + y = 0$ , then the equation of the other tangent through the origin is

1

3 x - y = 0

2

x + 3 y = 0

3

x - 3 y = 0

4

x + 2 y = 0

Explanation:

(C) : Given,
$3 x + y = 0$
Radius $= r$ (let)
The lengths of the perpendicular from the centre $(2, - 1)$ to the line $3 x + y = 0$
$r = \frac{6 - 1}{\sqrt{(9 + 1)}}$
$r = \frac{5}{\sqrt{10}}$
Let,
$y = m x$ be other tangent from origin,
Then, $\frac{| 2 m + 1 |}{\sqrt{(1 + m^{2})}} = \frac{5}{\sqrt{10}}$
On squaring both sides-
$\frac{4 m^{2} + 1 + 4 m}{1 + m^{2}} = \frac{25}{10}$
$8 m^{2} + 2 + 8 m = 5 + 5 m^{2}$
$3 m^{2} + 8 m - 3 = 0$
$m = - 3, 1 / 3$
As, -3 is the slope of given tangent so $1 / 3$ is that of the other tangent.
Hence,
$y = (\frac{1}{3}) x$
$3 y = x$
$3 y - x = 0$
$x - 3 y = 0$

WB JEE-2022

Application of Derivatives

85421 Consider the curve $y = b e^{- x / a}$ where $a$ and $b$ are non-zero real numbers. Then

1

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(0, 0)

2

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve, where the curve crosses the axis of

y

3

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(a, 0)

4

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(2 a, 0)

Explanation:

(B) : Given the curve,
$y = b e^{- x / a}$
On differentiating both sides w.r.t.x, we get-
$\frac{d y}{d x} = - \frac{b}{a} e^{\frac{- x}{a}}$
$∴ {(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a} e^{- 0 / a}$
${(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a}$
$∴$ Equation of tangent
$y - b = \frac{- b}{a} (x - 0) \Rightarrow y - b = \frac{- b x}{a}$
$\frac{y}{b} - 1 = \frac{- x}{a} \Rightarrow \frac{x}{a} + \frac{y}{b} = 1$
Hence, $\frac{x}{a} + \frac{y}{b} = 1$ is tangent to the curve where the curve crosses the axis of $y$ .

WB JEE-2020

Application of Derivatives

85422 If the tangent to the curve $y^{2} = x^{3} a t (m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{2})$ , then the value of $mM$ is

1

- \frac{1}{6}

2

- \frac{2}{9}

3

- \frac{1}{3}

4

- \frac{4}{9}

Explanation:

(D) : Given,
$y^{2} = x^{3}$
On differentiating both sides w.r.t.x, we get-
$2 y \frac{d y}{d x} = 3 x^{2}$
Slope, $m = \frac{3 x^{2}}{2 y}$
Slope of tangent at $(m^{2}, m^{3})$
$m_{1} = \frac{3 {(m^{2})}^{2}}{2 m^{3}} \Rightarrow m_{1} = \frac{3 m^{4}}{2 m^{3}}$
$m_{1} = \frac{3 m}{2}$
Slope at $[M^{2}, M^{3}]$
$m_{2} = \frac{3 M}{2}$
Now, $m_{1} m_{2} = - 1$
On putting the value of $m_{1} m_{2}$
$\frac{3 m}{2} \cdot \frac{3 M}{2} = - 1 \Rightarrow \frac{9 m M}{4} = - 1$
$m M = \frac{- 4}{9}$

Karnataka CET-2021

Application of Derivatives

85418 What is the $x$ -coordinate of the point on the curve $f (x) = \sqrt{x} (7 x - 6)$ , where the tangent is parallel to $x$ -axis

1

- \frac{1}{3}

2

\frac{2}{7}

3

\frac{6}{7}

4

\frac{1}{2}

Explanation:

(B) :Given,
$f (x) = \sqrt{x} (7 x - 6)$
$f (x) = 7 x^{3 / 2} - 6 x^{1 / 2}$
On differentiation both sides w.r.t.x, we get-
$f^{'} (x) = 7 \times \frac{3}{2} x^{1 / 2} - 6 \times \frac{1}{2} x^{- 1 / 2}$
When tangent is parallel to $x$ axis $f^{'} (x) = 0$
$\frac{21}{2} x^{1 / 2} - 3 x^{- 1 / 2} = 0 \Rightarrow \frac{21}{2} \sqrt{x} = \frac{3}{\sqrt{x}}$
$7 x = 2 \Rightarrow x = \frac{2}{7}$

BITSAT-2016

Application of Derivatives

85419 For the curve $x^{2} + 4 x y + 8 y^{2} = 64$ the tangents are parallel to the $x$ -axis only at the points

1

(0, 2 \sqrt{2})

and

(0, - 2 \sqrt{2})

2

(8, - 4)

and

(- 8, 4)

3

(8 \sqrt{2}, - 2 \sqrt{2})

and

(- 8 \sqrt{2}, 2 \sqrt{2})

4

(9, 0)

and

(- 8, 0)

Explanation:

(B) : Given,
$x^{2} + 4 x y + 8 y^{2} = 64$
On differentiating both sides w.r.t to $x$ we get -
$2 x + 4 (y + x \frac{d y}{d x}) + 16 y \frac{d y}{d x} = 0$
$2 x + 4 y + (4 x + 16 y) \frac{d y}{d x} = 0$
$\frac{d y}{d x} = - \frac{(x + 2 y)}{2 (x + 4 y)}$
Since, tangent are parallel to $x$ -axis only.
$\begin{aligned} \frac{d y}{d x} = 0 \Rightarrow 0 = - \frac{(x + 2 y)}{2 (x + 4 y)} \\ x + 2 y = 0 \end{aligned}$
On putting the values of $x$ from equation (i) in (ii), we get $4 y^{2} - 8 y^{2} + 8 y^{2} = 64$
$4 y^{2} = 64$
$y^{2} = 16$
$y = \pm 4$
On putting the value $y$ in equation (ii)
$x + 2 y = 0$
$x + 2 (4) = 0$
$x = - 8, y = 4$
$x = + 8, y = - 4$
Hence, required point are $(- 8, 4)$ and $(8, - 4)$

WB JEE-2013

Application of Derivatives

85420 If the equation of one tangent to the circle with centre at $(2, - 1)$ from the origin is $3 x + y = 0$ , then the equation of the other tangent through the origin is

1

3 x - y = 0

2

x + 3 y = 0

3

x - 3 y = 0

4

x + 2 y = 0

Explanation:

(C) : Given,
$3 x + y = 0$
Radius $= r$ (let)
The lengths of the perpendicular from the centre $(2, - 1)$ to the line $3 x + y = 0$
$r = \frac{6 - 1}{\sqrt{(9 + 1)}}$
$r = \frac{5}{\sqrt{10}}$
Let,
$y = m x$ be other tangent from origin,
Then, $\frac{| 2 m + 1 |}{\sqrt{(1 + m^{2})}} = \frac{5}{\sqrt{10}}$
On squaring both sides-
$\frac{4 m^{2} + 1 + 4 m}{1 + m^{2}} = \frac{25}{10}$
$8 m^{2} + 2 + 8 m = 5 + 5 m^{2}$
$3 m^{2} + 8 m - 3 = 0$
$m = - 3, 1 / 3$
As, -3 is the slope of given tangent so $1 / 3$ is that of the other tangent.
Hence,
$y = (\frac{1}{3}) x$
$3 y = x$
$3 y - x = 0$
$x - 3 y = 0$

WB JEE-2022

Application of Derivatives

85421 Consider the curve $y = b e^{- x / a}$ where $a$ and $b$ are non-zero real numbers. Then

1

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(0, 0)

2

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve, where the curve crosses the axis of

y

3

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(a, 0)

4

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(2 a, 0)

Explanation:

(B) : Given the curve,
$y = b e^{- x / a}$
On differentiating both sides w.r.t.x, we get-
$\frac{d y}{d x} = - \frac{b}{a} e^{\frac{- x}{a}}$
$∴ {(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a} e^{- 0 / a}$
${(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a}$
$∴$ Equation of tangent
$y - b = \frac{- b}{a} (x - 0) \Rightarrow y - b = \frac{- b x}{a}$
$\frac{y}{b} - 1 = \frac{- x}{a} \Rightarrow \frac{x}{a} + \frac{y}{b} = 1$
Hence, $\frac{x}{a} + \frac{y}{b} = 1$ is tangent to the curve where the curve crosses the axis of $y$ .

WB JEE-2020

Application of Derivatives

85422 If the tangent to the curve $y^{2} = x^{3} a t (m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{2})$ , then the value of $mM$ is

1

- \frac{1}{6}

2

- \frac{2}{9}

3

- \frac{1}{3}

4

- \frac{4}{9}

Explanation:

(D) : Given,
$y^{2} = x^{3}$
On differentiating both sides w.r.t.x, we get-
$2 y \frac{d y}{d x} = 3 x^{2}$
Slope, $m = \frac{3 x^{2}}{2 y}$
Slope of tangent at $(m^{2}, m^{3})$
$m_{1} = \frac{3 {(m^{2})}^{2}}{2 m^{3}} \Rightarrow m_{1} = \frac{3 m^{4}}{2 m^{3}}$
$m_{1} = \frac{3 m}{2}$
Slope at $[M^{2}, M^{3}]$
$m_{2} = \frac{3 M}{2}$
Now, $m_{1} m_{2} = - 1$
On putting the value of $m_{1} m_{2}$
$\frac{3 m}{2} \cdot \frac{3 M}{2} = - 1 \Rightarrow \frac{9 m M}{4} = - 1$
$m M = \frac{- 4}{9}$

Karnataka CET-2021

Application of Derivatives

85418 What is the $x$ -coordinate of the point on the curve $f (x) = \sqrt{x} (7 x - 6)$ , where the tangent is parallel to $x$ -axis

1

- \frac{1}{3}

2

\frac{2}{7}

3

\frac{6}{7}

4

\frac{1}{2}

Explanation:

(B) :Given,
$f (x) = \sqrt{x} (7 x - 6)$
$f (x) = 7 x^{3 / 2} - 6 x^{1 / 2}$
On differentiation both sides w.r.t.x, we get-
$f^{'} (x) = 7 \times \frac{3}{2} x^{1 / 2} - 6 \times \frac{1}{2} x^{- 1 / 2}$
When tangent is parallel to $x$ axis $f^{'} (x) = 0$
$\frac{21}{2} x^{1 / 2} - 3 x^{- 1 / 2} = 0 \Rightarrow \frac{21}{2} \sqrt{x} = \frac{3}{\sqrt{x}}$
$7 x = 2 \Rightarrow x = \frac{2}{7}$

BITSAT-2016

Application of Derivatives

85419 For the curve $x^{2} + 4 x y + 8 y^{2} = 64$ the tangents are parallel to the $x$ -axis only at the points

1

(0, 2 \sqrt{2})

and

(0, - 2 \sqrt{2})

2

(8, - 4)

and

(- 8, 4)

3

(8 \sqrt{2}, - 2 \sqrt{2})

and

(- 8 \sqrt{2}, 2 \sqrt{2})

4

(9, 0)

and

(- 8, 0)

Explanation:

(B) : Given,
$x^{2} + 4 x y + 8 y^{2} = 64$
On differentiating both sides w.r.t to $x$ we get -
$2 x + 4 (y + x \frac{d y}{d x}) + 16 y \frac{d y}{d x} = 0$
$2 x + 4 y + (4 x + 16 y) \frac{d y}{d x} = 0$
$\frac{d y}{d x} = - \frac{(x + 2 y)}{2 (x + 4 y)}$
Since, tangent are parallel to $x$ -axis only.
$\begin{aligned} \frac{d y}{d x} = 0 \Rightarrow 0 = - \frac{(x + 2 y)}{2 (x + 4 y)} \\ x + 2 y = 0 \end{aligned}$
On putting the values of $x$ from equation (i) in (ii), we get $4 y^{2} - 8 y^{2} + 8 y^{2} = 64$
$4 y^{2} = 64$
$y^{2} = 16$
$y = \pm 4$
On putting the value $y$ in equation (ii)
$x + 2 y = 0$
$x + 2 (4) = 0$
$x = - 8, y = 4$
$x = + 8, y = - 4$
Hence, required point are $(- 8, 4)$ and $(8, - 4)$

WB JEE-2013

Application of Derivatives

85420 If the equation of one tangent to the circle with centre at $(2, - 1)$ from the origin is $3 x + y = 0$ , then the equation of the other tangent through the origin is

1

3 x - y = 0

2

x + 3 y = 0

3

x - 3 y = 0

4

x + 2 y = 0

Explanation:

(C) : Given,
$3 x + y = 0$
Radius $= r$ (let)
The lengths of the perpendicular from the centre $(2, - 1)$ to the line $3 x + y = 0$
$r = \frac{6 - 1}{\sqrt{(9 + 1)}}$
$r = \frac{5}{\sqrt{10}}$
Let,
$y = m x$ be other tangent from origin,
Then, $\frac{| 2 m + 1 |}{\sqrt{(1 + m^{2})}} = \frac{5}{\sqrt{10}}$
On squaring both sides-
$\frac{4 m^{2} + 1 + 4 m}{1 + m^{2}} = \frac{25}{10}$
$8 m^{2} + 2 + 8 m = 5 + 5 m^{2}$
$3 m^{2} + 8 m - 3 = 0$
$m = - 3, 1 / 3$
As, -3 is the slope of given tangent so $1 / 3$ is that of the other tangent.
Hence,
$y = (\frac{1}{3}) x$
$3 y = x$
$3 y - x = 0$
$x - 3 y = 0$

WB JEE-2022

Application of Derivatives

85421 Consider the curve $y = b e^{- x / a}$ where $a$ and $b$ are non-zero real numbers. Then

1

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(0, 0)

2

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve, where the curve crosses the axis of

y

3

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(a, 0)

4

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(2 a, 0)

Explanation:

(B) : Given the curve,
$y = b e^{- x / a}$
On differentiating both sides w.r.t.x, we get-
$\frac{d y}{d x} = - \frac{b}{a} e^{\frac{- x}{a}}$
$∴ {(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a} e^{- 0 / a}$
${(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a}$
$∴$ Equation of tangent
$y - b = \frac{- b}{a} (x - 0) \Rightarrow y - b = \frac{- b x}{a}$
$\frac{y}{b} - 1 = \frac{- x}{a} \Rightarrow \frac{x}{a} + \frac{y}{b} = 1$
Hence, $\frac{x}{a} + \frac{y}{b} = 1$ is tangent to the curve where the curve crosses the axis of $y$ .

WB JEE-2020

Application of Derivatives

85422 If the tangent to the curve $y^{2} = x^{3} a t (m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{2})$ , then the value of $mM$ is

1

- \frac{1}{6}

2

- \frac{2}{9}

3

- \frac{1}{3}

4

- \frac{4}{9}

Explanation:

(D) : Given,
$y^{2} = x^{3}$
On differentiating both sides w.r.t.x, we get-
$2 y \frac{d y}{d x} = 3 x^{2}$
Slope, $m = \frac{3 x^{2}}{2 y}$
Slope of tangent at $(m^{2}, m^{3})$
$m_{1} = \frac{3 {(m^{2})}^{2}}{2 m^{3}} \Rightarrow m_{1} = \frac{3 m^{4}}{2 m^{3}}$
$m_{1} = \frac{3 m}{2}$
Slope at $[M^{2}, M^{3}]$
$m_{2} = \frac{3 M}{2}$
Now, $m_{1} m_{2} = - 1$
On putting the value of $m_{1} m_{2}$
$\frac{3 m}{2} \cdot \frac{3 M}{2} = - 1 \Rightarrow \frac{9 m M}{4} = - 1$
$m M = \frac{- 4}{9}$

Karnataka CET-2021

Application of Derivatives

85418 What is the $x$ -coordinate of the point on the curve $f (x) = \sqrt{x} (7 x - 6)$ , where the tangent is parallel to $x$ -axis

1

- \frac{1}{3}

2

\frac{2}{7}

3

\frac{6}{7}

4

\frac{1}{2}

Explanation:

(B) :Given,
$f (x) = \sqrt{x} (7 x - 6)$
$f (x) = 7 x^{3 / 2} - 6 x^{1 / 2}$
On differentiation both sides w.r.t.x, we get-
$f^{'} (x) = 7 \times \frac{3}{2} x^{1 / 2} - 6 \times \frac{1}{2} x^{- 1 / 2}$
When tangent is parallel to $x$ axis $f^{'} (x) = 0$
$\frac{21}{2} x^{1 / 2} - 3 x^{- 1 / 2} = 0 \Rightarrow \frac{21}{2} \sqrt{x} = \frac{3}{\sqrt{x}}$
$7 x = 2 \Rightarrow x = \frac{2}{7}$

BITSAT-2016

Application of Derivatives

85419 For the curve $x^{2} + 4 x y + 8 y^{2} = 64$ the tangents are parallel to the $x$ -axis only at the points

1

(0, 2 \sqrt{2})

and

(0, - 2 \sqrt{2})

2

(8, - 4)

and

(- 8, 4)

3

(8 \sqrt{2}, - 2 \sqrt{2})

and

(- 8 \sqrt{2}, 2 \sqrt{2})

4

(9, 0)

and

(- 8, 0)

Explanation:

(B) : Given,
$x^{2} + 4 x y + 8 y^{2} = 64$
On differentiating both sides w.r.t to $x$ we get -
$2 x + 4 (y + x \frac{d y}{d x}) + 16 y \frac{d y}{d x} = 0$
$2 x + 4 y + (4 x + 16 y) \frac{d y}{d x} = 0$
$\frac{d y}{d x} = - \frac{(x + 2 y)}{2 (x + 4 y)}$
Since, tangent are parallel to $x$ -axis only.
$\begin{aligned} \frac{d y}{d x} = 0 \Rightarrow 0 = - \frac{(x + 2 y)}{2 (x + 4 y)} \\ x + 2 y = 0 \end{aligned}$
On putting the values of $x$ from equation (i) in (ii), we get $4 y^{2} - 8 y^{2} + 8 y^{2} = 64$
$4 y^{2} = 64$
$y^{2} = 16$
$y = \pm 4$
On putting the value $y$ in equation (ii)
$x + 2 y = 0$
$x + 2 (4) = 0$
$x = - 8, y = 4$
$x = + 8, y = - 4$
Hence, required point are $(- 8, 4)$ and $(8, - 4)$

WB JEE-2013

Application of Derivatives

85420 If the equation of one tangent to the circle with centre at $(2, - 1)$ from the origin is $3 x + y = 0$ , then the equation of the other tangent through the origin is

1

3 x - y = 0

2

x + 3 y = 0

3

x - 3 y = 0

4

x + 2 y = 0

Explanation:

(C) : Given,
$3 x + y = 0$
Radius $= r$ (let)
The lengths of the perpendicular from the centre $(2, - 1)$ to the line $3 x + y = 0$
$r = \frac{6 - 1}{\sqrt{(9 + 1)}}$
$r = \frac{5}{\sqrt{10}}$
Let,
$y = m x$ be other tangent from origin,
Then, $\frac{| 2 m + 1 |}{\sqrt{(1 + m^{2})}} = \frac{5}{\sqrt{10}}$
On squaring both sides-
$\frac{4 m^{2} + 1 + 4 m}{1 + m^{2}} = \frac{25}{10}$
$8 m^{2} + 2 + 8 m = 5 + 5 m^{2}$
$3 m^{2} + 8 m - 3 = 0$
$m = - 3, 1 / 3$
As, -3 is the slope of given tangent so $1 / 3$ is that of the other tangent.
Hence,
$y = (\frac{1}{3}) x$
$3 y = x$
$3 y - x = 0$
$x - 3 y = 0$

WB JEE-2022

Application of Derivatives

85421 Consider the curve $y = b e^{- x / a}$ where $a$ and $b$ are non-zero real numbers. Then

1

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(0, 0)

2

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve, where the curve crosses the axis of

y

3

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(a, 0)

4

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(2 a, 0)

Explanation:

(B) : Given the curve,
$y = b e^{- x / a}$
On differentiating both sides w.r.t.x, we get-
$\frac{d y}{d x} = - \frac{b}{a} e^{\frac{- x}{a}}$
$∴ {(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a} e^{- 0 / a}$
${(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a}$
$∴$ Equation of tangent
$y - b = \frac{- b}{a} (x - 0) \Rightarrow y - b = \frac{- b x}{a}$
$\frac{y}{b} - 1 = \frac{- x}{a} \Rightarrow \frac{x}{a} + \frac{y}{b} = 1$
Hence, $\frac{x}{a} + \frac{y}{b} = 1$ is tangent to the curve where the curve crosses the axis of $y$ .

WB JEE-2020

Application of Derivatives

85422 If the tangent to the curve $y^{2} = x^{3} a t (m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{2})$ , then the value of $mM$ is

1

- \frac{1}{6}

2

- \frac{2}{9}

3

- \frac{1}{3}

4

- \frac{4}{9}

Explanation:

(D) : Given,
$y^{2} = x^{3}$
On differentiating both sides w.r.t.x, we get-
$2 y \frac{d y}{d x} = 3 x^{2}$
Slope, $m = \frac{3 x^{2}}{2 y}$
Slope of tangent at $(m^{2}, m^{3})$
$m_{1} = \frac{3 {(m^{2})}^{2}}{2 m^{3}} \Rightarrow m_{1} = \frac{3 m^{4}}{2 m^{3}}$
$m_{1} = \frac{3 m}{2}$
Slope at $[M^{2}, M^{3}]$
$m_{2} = \frac{3 M}{2}$
Now, $m_{1} m_{2} = - 1$
On putting the value of $m_{1} m_{2}$
$\frac{3 m}{2} \cdot \frac{3 M}{2} = - 1 \Rightarrow \frac{9 m M}{4} = - 1$
$m M = \frac{- 4}{9}$

Karnataka CET-2021

Application of Derivatives

85418 What is the $x$ -coordinate of the point on the curve $f (x) = \sqrt{x} (7 x - 6)$ , where the tangent is parallel to $x$ -axis

1

- \frac{1}{3}

2

\frac{2}{7}

3

\frac{6}{7}

4

\frac{1}{2}

Explanation:

(B) :Given,
$f (x) = \sqrt{x} (7 x - 6)$
$f (x) = 7 x^{3 / 2} - 6 x^{1 / 2}$
On differentiation both sides w.r.t.x, we get-
$f^{'} (x) = 7 \times \frac{3}{2} x^{1 / 2} - 6 \times \frac{1}{2} x^{- 1 / 2}$
When tangent is parallel to $x$ axis $f^{'} (x) = 0$
$\frac{21}{2} x^{1 / 2} - 3 x^{- 1 / 2} = 0 \Rightarrow \frac{21}{2} \sqrt{x} = \frac{3}{\sqrt{x}}$
$7 x = 2 \Rightarrow x = \frac{2}{7}$

BITSAT-2016

Application of Derivatives

85419 For the curve $x^{2} + 4 x y + 8 y^{2} = 64$ the tangents are parallel to the $x$ -axis only at the points

1

(0, 2 \sqrt{2})

and

(0, - 2 \sqrt{2})

2

(8, - 4)

and

(- 8, 4)

3

(8 \sqrt{2}, - 2 \sqrt{2})

and

(- 8 \sqrt{2}, 2 \sqrt{2})

4

(9, 0)

and

(- 8, 0)

Explanation:

(B) : Given,
$x^{2} + 4 x y + 8 y^{2} = 64$
On differentiating both sides w.r.t to $x$ we get -
$2 x + 4 (y + x \frac{d y}{d x}) + 16 y \frac{d y}{d x} = 0$
$2 x + 4 y + (4 x + 16 y) \frac{d y}{d x} = 0$
$\frac{d y}{d x} = - \frac{(x + 2 y)}{2 (x + 4 y)}$
Since, tangent are parallel to $x$ -axis only.
$\begin{aligned} \frac{d y}{d x} = 0 \Rightarrow 0 = - \frac{(x + 2 y)}{2 (x + 4 y)} \\ x + 2 y = 0 \end{aligned}$
On putting the values of $x$ from equation (i) in (ii), we get $4 y^{2} - 8 y^{2} + 8 y^{2} = 64$
$4 y^{2} = 64$
$y^{2} = 16$
$y = \pm 4$
On putting the value $y$ in equation (ii)
$x + 2 y = 0$
$x + 2 (4) = 0$
$x = - 8, y = 4$
$x = + 8, y = - 4$
Hence, required point are $(- 8, 4)$ and $(8, - 4)$

WB JEE-2013

Application of Derivatives

85420 If the equation of one tangent to the circle with centre at $(2, - 1)$ from the origin is $3 x + y = 0$ , then the equation of the other tangent through the origin is

1

3 x - y = 0

2

x + 3 y = 0

3

x - 3 y = 0

4

x + 2 y = 0

Explanation:

(C) : Given,
$3 x + y = 0$
Radius $= r$ (let)
The lengths of the perpendicular from the centre $(2, - 1)$ to the line $3 x + y = 0$
$r = \frac{6 - 1}{\sqrt{(9 + 1)}}$
$r = \frac{5}{\sqrt{10}}$
Let,
$y = m x$ be other tangent from origin,
Then, $\frac{| 2 m + 1 |}{\sqrt{(1 + m^{2})}} = \frac{5}{\sqrt{10}}$
On squaring both sides-
$\frac{4 m^{2} + 1 + 4 m}{1 + m^{2}} = \frac{25}{10}$
$8 m^{2} + 2 + 8 m = 5 + 5 m^{2}$
$3 m^{2} + 8 m - 3 = 0$
$m = - 3, 1 / 3$
As, -3 is the slope of given tangent so $1 / 3$ is that of the other tangent.
Hence,
$y = (\frac{1}{3}) x$
$3 y = x$
$3 y - x = 0$
$x - 3 y = 0$

WB JEE-2022

Application of Derivatives

85421 Consider the curve $y = b e^{- x / a}$ where $a$ and $b$ are non-zero real numbers. Then

1

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(0, 0)

2

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve, where the curve crosses the axis of

y

3

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(a, 0)

4

\frac{x}{a} + \frac{y}{b} = 1

is tangent to the curve at

(2 a, 0)

Explanation:

(B) : Given the curve,
$y = b e^{- x / a}$
On differentiating both sides w.r.t.x, we get-
$\frac{d y}{d x} = - \frac{b}{a} e^{\frac{- x}{a}}$
$∴ {(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a} e^{- 0 / a}$
${(\frac{dy}{dx})}_{(0, b)} = \frac{- b}{a}$
$∴$ Equation of tangent
$y - b = \frac{- b}{a} (x - 0) \Rightarrow y - b = \frac{- b x}{a}$
$\frac{y}{b} - 1 = \frac{- x}{a} \Rightarrow \frac{x}{a} + \frac{y}{b} = 1$
Hence, $\frac{x}{a} + \frac{y}{b} = 1$ is tangent to the curve where the curve crosses the axis of $y$ .

WB JEE-2020

Application of Derivatives

85422 If the tangent to the curve $y^{2} = x^{3} a t (m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{2})$ , then the value of $mM$ is

1

- \frac{1}{6}

2

- \frac{2}{9}

3

- \frac{1}{3}

4

- \frac{4}{9}

Explanation:

(D) : Given,
$y^{2} = x^{3}$
On differentiating both sides w.r.t.x, we get-
$2 y \frac{d y}{d x} = 3 x^{2}$
Slope, $m = \frac{3 x^{2}}{2 y}$
Slope of tangent at $(m^{2}, m^{3})$
$m_{1} = \frac{3 {(m^{2})}^{2}}{2 m^{3}} \Rightarrow m_{1} = \frac{3 m^{4}}{2 m^{3}}$
$m_{1} = \frac{3 m}{2}$
Slope at $[M^{2}, M^{3}]$
$m_{2} = \frac{3 M}{2}$
Now, $m_{1} m_{2} = - 1$
On putting the value of $m_{1} m_{2}$
$\frac{3 m}{2} \cdot \frac{3 M}{2} = - 1 \Rightarrow \frac{9 m M}{4} = - 1$
$m M = \frac{- 4}{9}$

Karnataka CET-2021

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