QBManager

Parabola

120100 The equation $y^{2} + 4 x + 4 y + k = 0$ represents a parabola whose latusrectum is

1 1

2 2

3 3

4 4

Explanation:

D Given,
$y^{2} + 4 x + 4 y + k = 0$
$y^{2} + 4 y + (2)^{2} - (2)^{2} + 4 x + k = 0$
$(y + 2)^{2} = - 4 x - k + 4$
$(y + 2)^{2} = - 4 (x + \frac{k}{4} - 1)$
$(y + 2)^{2} = - 4 (x + \frac{k - 4}{4})$
Now, compare with,
${(y - y_{1})}^{2} = - 4 a (x - x_{1})$ Hence, latusrectum is $4 a = 4 \times 1 = 4$

WB JEE-2012

Parabola

120101 The two lines $t y = x + t^{2}$ and $y + t x = 2 t + t^{3}$ intersect at the point lines on the curve whose equation is

1

y^{2} = 4 x

2

y^{2} = - 4 x

3

x^{2} = 4 y

4

x^{2} = - 4 y

Explanation:

A Given,
ty $= x + t^{2}$
and $y + t x = 2 t + t^{3}$
From equation (i) and (ii) we get-
$y + t (t y - t^{2}) = 2 t + t^{3}$
$y (1 + t^{2}) - t^{3} = 2 t + t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y = \frac{2 t (1 + t^{2})}{(1 + t^{2})}$
$y = 2 t \Rightarrow y^{2} = 4 t^{2}$
Then, $x = t^{2} \Rightarrow y^{2} = 4 x$
So, the point of intersection lies on the curve $y^{2} = 4 x$

AMU-2016

Parabola

120102 The point on the parabola $y^{2} = 64 x$ which is nearest to the line $4 x + 3 y + 35 = 0$ has coordinates

1

(9, - 24)

2

(1, - 81)

3

(4, - 16)

4

(- 9, - 24)

Explanation:

A :Given, $y^{2} = 64 x$
Differentiate, w.r.t. ' $x$ '
$2 y \frac{d y}{d x} = 64$
$\frac{d y}{d x} = \frac{32}{y}$
Also,
$4 x + 3 y + 35 = 0$
$y = - \frac{4 x}{3} - \frac{35}{3}$ Hence, slope $(m) = \frac{- 4}{3} = \frac{dy}{dx}$
$\frac{- 4}{3} = \frac{32}{y}$
$y = - 24$
$∴ y^{2} = 64 x$
$x = \frac{y^{2}}{64} = \frac{(- 24)^{2}}{64}$
$x = 9$
Hence, required point is $= (9, - 24)$

WB JEE-2014

Parabola

120103 The value of $λ$ for which the curve
$(7 x + 5)^{2} + (7 y + 3)^{2} = λ^{2} (4 x + 3 y - 24)^{2}$
represents a parabola is

1

\pm \frac{6}{5}

2

\pm \frac{7}{5}

3

\pm \frac{1}{5}

4

\pm \frac{2}{5}

Explanation:

B Given,
$(7 x + 5)^{2} + (7 y + 3)^{3} = λ^{2} (4 x + 3 y - 24)^{2}$
$49 x^{2} + 25 + 70 x + 49 y^{2} + 9 + 42 y = λ^{2} (16 x^{2} + 9 y^{2} +$
$576 + 24 xy - 144 y - 192 x)$
$(49 - 16 λ)^{2} x^{2} + (49 - 9 λ^{2}) y^{2} + (70 + 192 λ^{2}) x + (42 +$
$144 λ) y - 24 λ^{2} xy + (25 - 576 λ^{2}) = 0$
$If, ax^{2} + 2 hxy + by y^{2} + 2 yx + 2 fy + c = 0$
$Then, for parabola, ab = h^{2}$
$Hence, (49 - 16 λ^{2}) (49 - 9 λ^{2}) = {(- 12 λ^{2})}^{2}$
$2401 - 441 λ^{2} - 784 λ^{2} + 144 λ^{4} = 144 λ^{4}$
$2401 - 1225 λ^{2} = 0$
$λ^{2} = \frac{2401}{1225}$
$λ = \pm \frac{7}{5}$

WB JEE-2014

Parabola

120104 A line passing through the point of intersection of $x + y = 4$ and $x - y = 2$ makes an angle $\tan^{- 1}$ $(\frac{3}{2})$ with the $X$ -axis. It intersects the parabola $y^{2} = 4 (x - 3)$ at points $(x_{1} - x_{2})$ and $(x_{2}, y_{2})$ respectively. Then, $| x_{1} - x_{2} |$ is equal to

1

\frac{16}{9}

2

\frac{32}{9}

3

\frac{40}{9}

4

\frac{80}{9}

Explanation:

B Given,
Lines: $x + y = 4 & x - y = 2$
$y = 4 - x & y = x - 2$
$∴ 4 - x = x - 2$
$2 x = 6$
$x = 3$ and $y = 1$
Now, equation of line is $(3, 1)$ and slope $(\frac{3}{4})$
$(y - 1) = \frac{3}{4} (x - 3)$
$3 x - 4 y = 5$
Also, $y^{2} = 4 (x - 3)$
${(\frac{3 x - 5}{4})}^{2} = 4 (x - 3)$
$9 x^{2} + 25 - 30 x = 64 x - 192$
$9 x^{2} - 94 x + 217 = 0$
$x = 7, \frac{31}{9}$
Hence, $| x_{1} - x_{2} | = | 7 - \frac{31}{9} |$
$| x_{1} - x_{2} | = \frac{32}{9}$

WB JEE-2016

Parabola

120100 The equation $y^{2} + 4 x + 4 y + k = 0$ represents a parabola whose latusrectum is

1 1

2 2

3 3

4 4

Explanation:

D Given,
$y^{2} + 4 x + 4 y + k = 0$
$y^{2} + 4 y + (2)^{2} - (2)^{2} + 4 x + k = 0$
$(y + 2)^{2} = - 4 x - k + 4$
$(y + 2)^{2} = - 4 (x + \frac{k}{4} - 1)$
$(y + 2)^{2} = - 4 (x + \frac{k - 4}{4})$
Now, compare with,
${(y - y_{1})}^{2} = - 4 a (x - x_{1})$ Hence, latusrectum is $4 a = 4 \times 1 = 4$

WB JEE-2012

Parabola

120101 The two lines $t y = x + t^{2}$ and $y + t x = 2 t + t^{3}$ intersect at the point lines on the curve whose equation is

1

y^{2} = 4 x

2

y^{2} = - 4 x

3

x^{2} = 4 y

4

x^{2} = - 4 y

Explanation:

A Given,
ty $= x + t^{2}$
and $y + t x = 2 t + t^{3}$
From equation (i) and (ii) we get-
$y + t (t y - t^{2}) = 2 t + t^{3}$
$y (1 + t^{2}) - t^{3} = 2 t + t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y = \frac{2 t (1 + t^{2})}{(1 + t^{2})}$
$y = 2 t \Rightarrow y^{2} = 4 t^{2}$
Then, $x = t^{2} \Rightarrow y^{2} = 4 x$
So, the point of intersection lies on the curve $y^{2} = 4 x$

AMU-2016

Parabola

120102 The point on the parabola $y^{2} = 64 x$ which is nearest to the line $4 x + 3 y + 35 = 0$ has coordinates

1

(9, - 24)

2

(1, - 81)

3

(4, - 16)

4

(- 9, - 24)

Explanation:

A :Given, $y^{2} = 64 x$
Differentiate, w.r.t. ' $x$ '
$2 y \frac{d y}{d x} = 64$
$\frac{d y}{d x} = \frac{32}{y}$
Also,
$4 x + 3 y + 35 = 0$
$y = - \frac{4 x}{3} - \frac{35}{3}$ Hence, slope $(m) = \frac{- 4}{3} = \frac{dy}{dx}$
$\frac{- 4}{3} = \frac{32}{y}$
$y = - 24$
$∴ y^{2} = 64 x$
$x = \frac{y^{2}}{64} = \frac{(- 24)^{2}}{64}$
$x = 9$
Hence, required point is $= (9, - 24)$

WB JEE-2014

Parabola

120103 The value of $λ$ for which the curve
$(7 x + 5)^{2} + (7 y + 3)^{2} = λ^{2} (4 x + 3 y - 24)^{2}$
represents a parabola is

1

\pm \frac{6}{5}

2

\pm \frac{7}{5}

3

\pm \frac{1}{5}

4

\pm \frac{2}{5}

Explanation:

B Given,
$(7 x + 5)^{2} + (7 y + 3)^{3} = λ^{2} (4 x + 3 y - 24)^{2}$
$49 x^{2} + 25 + 70 x + 49 y^{2} + 9 + 42 y = λ^{2} (16 x^{2} + 9 y^{2} +$
$576 + 24 xy - 144 y - 192 x)$
$(49 - 16 λ)^{2} x^{2} + (49 - 9 λ^{2}) y^{2} + (70 + 192 λ^{2}) x + (42 +$
$144 λ) y - 24 λ^{2} xy + (25 - 576 λ^{2}) = 0$
$If, ax^{2} + 2 hxy + by y^{2} + 2 yx + 2 fy + c = 0$
$Then, for parabola, ab = h^{2}$
$Hence, (49 - 16 λ^{2}) (49 - 9 λ^{2}) = {(- 12 λ^{2})}^{2}$
$2401 - 441 λ^{2} - 784 λ^{2} + 144 λ^{4} = 144 λ^{4}$
$2401 - 1225 λ^{2} = 0$
$λ^{2} = \frac{2401}{1225}$
$λ = \pm \frac{7}{5}$

WB JEE-2014

Parabola

120104 A line passing through the point of intersection of $x + y = 4$ and $x - y = 2$ makes an angle $\tan^{- 1}$ $(\frac{3}{2})$ with the $X$ -axis. It intersects the parabola $y^{2} = 4 (x - 3)$ at points $(x_{1} - x_{2})$ and $(x_{2}, y_{2})$ respectively. Then, $| x_{1} - x_{2} |$ is equal to

1

\frac{16}{9}

2

\frac{32}{9}

3

\frac{40}{9}

4

\frac{80}{9}

Explanation:

B Given,
Lines: $x + y = 4 & x - y = 2$
$y = 4 - x & y = x - 2$
$∴ 4 - x = x - 2$
$2 x = 6$
$x = 3$ and $y = 1$
Now, equation of line is $(3, 1)$ and slope $(\frac{3}{4})$
$(y - 1) = \frac{3}{4} (x - 3)$
$3 x - 4 y = 5$
Also, $y^{2} = 4 (x - 3)$
${(\frac{3 x - 5}{4})}^{2} = 4 (x - 3)$
$9 x^{2} + 25 - 30 x = 64 x - 192$
$9 x^{2} - 94 x + 217 = 0$
$x = 7, \frac{31}{9}$
Hence, $| x_{1} - x_{2} | = | 7 - \frac{31}{9} |$
$| x_{1} - x_{2} | = \frac{32}{9}$

WB JEE-2016

Parabola

120100 The equation $y^{2} + 4 x + 4 y + k = 0$ represents a parabola whose latusrectum is

1 1

2 2

3 3

4 4

Explanation:

D Given,
$y^{2} + 4 x + 4 y + k = 0$
$y^{2} + 4 y + (2)^{2} - (2)^{2} + 4 x + k = 0$
$(y + 2)^{2} = - 4 x - k + 4$
$(y + 2)^{2} = - 4 (x + \frac{k}{4} - 1)$
$(y + 2)^{2} = - 4 (x + \frac{k - 4}{4})$
Now, compare with,
${(y - y_{1})}^{2} = - 4 a (x - x_{1})$ Hence, latusrectum is $4 a = 4 \times 1 = 4$

WB JEE-2012

Parabola

120101 The two lines $t y = x + t^{2}$ and $y + t x = 2 t + t^{3}$ intersect at the point lines on the curve whose equation is

1

y^{2} = 4 x

2

y^{2} = - 4 x

3

x^{2} = 4 y

4

x^{2} = - 4 y

Explanation:

A Given,
ty $= x + t^{2}$
and $y + t x = 2 t + t^{3}$
From equation (i) and (ii) we get-
$y + t (t y - t^{2}) = 2 t + t^{3}$
$y (1 + t^{2}) - t^{3} = 2 t + t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y = \frac{2 t (1 + t^{2})}{(1 + t^{2})}$
$y = 2 t \Rightarrow y^{2} = 4 t^{2}$
Then, $x = t^{2} \Rightarrow y^{2} = 4 x$
So, the point of intersection lies on the curve $y^{2} = 4 x$

AMU-2016

Parabola

120102 The point on the parabola $y^{2} = 64 x$ which is nearest to the line $4 x + 3 y + 35 = 0$ has coordinates

1

(9, - 24)

2

(1, - 81)

3

(4, - 16)

4

(- 9, - 24)

Explanation:

A :Given, $y^{2} = 64 x$
Differentiate, w.r.t. ' $x$ '
$2 y \frac{d y}{d x} = 64$
$\frac{d y}{d x} = \frac{32}{y}$
Also,
$4 x + 3 y + 35 = 0$
$y = - \frac{4 x}{3} - \frac{35}{3}$ Hence, slope $(m) = \frac{- 4}{3} = \frac{dy}{dx}$
$\frac{- 4}{3} = \frac{32}{y}$
$y = - 24$
$∴ y^{2} = 64 x$
$x = \frac{y^{2}}{64} = \frac{(- 24)^{2}}{64}$
$x = 9$
Hence, required point is $= (9, - 24)$

WB JEE-2014

Parabola

120103 The value of $λ$ for which the curve
$(7 x + 5)^{2} + (7 y + 3)^{2} = λ^{2} (4 x + 3 y - 24)^{2}$
represents a parabola is

1

\pm \frac{6}{5}

2

\pm \frac{7}{5}

3

\pm \frac{1}{5}

4

\pm \frac{2}{5}

Explanation:

B Given,
$(7 x + 5)^{2} + (7 y + 3)^{3} = λ^{2} (4 x + 3 y - 24)^{2}$
$49 x^{2} + 25 + 70 x + 49 y^{2} + 9 + 42 y = λ^{2} (16 x^{2} + 9 y^{2} +$
$576 + 24 xy - 144 y - 192 x)$
$(49 - 16 λ)^{2} x^{2} + (49 - 9 λ^{2}) y^{2} + (70 + 192 λ^{2}) x + (42 +$
$144 λ) y - 24 λ^{2} xy + (25 - 576 λ^{2}) = 0$
$If, ax^{2} + 2 hxy + by y^{2} + 2 yx + 2 fy + c = 0$
$Then, for parabola, ab = h^{2}$
$Hence, (49 - 16 λ^{2}) (49 - 9 λ^{2}) = {(- 12 λ^{2})}^{2}$
$2401 - 441 λ^{2} - 784 λ^{2} + 144 λ^{4} = 144 λ^{4}$
$2401 - 1225 λ^{2} = 0$
$λ^{2} = \frac{2401}{1225}$
$λ = \pm \frac{7}{5}$

WB JEE-2014

Parabola

120104 A line passing through the point of intersection of $x + y = 4$ and $x - y = 2$ makes an angle $\tan^{- 1}$ $(\frac{3}{2})$ with the $X$ -axis. It intersects the parabola $y^{2} = 4 (x - 3)$ at points $(x_{1} - x_{2})$ and $(x_{2}, y_{2})$ respectively. Then, $| x_{1} - x_{2} |$ is equal to

1

\frac{16}{9}

2

\frac{32}{9}

3

\frac{40}{9}

4

\frac{80}{9}

Explanation:

B Given,
Lines: $x + y = 4 & x - y = 2$
$y = 4 - x & y = x - 2$
$∴ 4 - x = x - 2$
$2 x = 6$
$x = 3$ and $y = 1$
Now, equation of line is $(3, 1)$ and slope $(\frac{3}{4})$
$(y - 1) = \frac{3}{4} (x - 3)$
$3 x - 4 y = 5$
Also, $y^{2} = 4 (x - 3)$
${(\frac{3 x - 5}{4})}^{2} = 4 (x - 3)$
$9 x^{2} + 25 - 30 x = 64 x - 192$
$9 x^{2} - 94 x + 217 = 0$
$x = 7, \frac{31}{9}$
Hence, $| x_{1} - x_{2} | = | 7 - \frac{31}{9} |$
$| x_{1} - x_{2} | = \frac{32}{9}$

WB JEE-2016

Parabola

120100 The equation $y^{2} + 4 x + 4 y + k = 0$ represents a parabola whose latusrectum is

1 1

2 2

3 3

4 4

Explanation:

D Given,
$y^{2} + 4 x + 4 y + k = 0$
$y^{2} + 4 y + (2)^{2} - (2)^{2} + 4 x + k = 0$
$(y + 2)^{2} = - 4 x - k + 4$
$(y + 2)^{2} = - 4 (x + \frac{k}{4} - 1)$
$(y + 2)^{2} = - 4 (x + \frac{k - 4}{4})$
Now, compare with,
${(y - y_{1})}^{2} = - 4 a (x - x_{1})$ Hence, latusrectum is $4 a = 4 \times 1 = 4$

WB JEE-2012

Parabola

120101 The two lines $t y = x + t^{2}$ and $y + t x = 2 t + t^{3}$ intersect at the point lines on the curve whose equation is

1

y^{2} = 4 x

2

y^{2} = - 4 x

3

x^{2} = 4 y

4

x^{2} = - 4 y

Explanation:

A Given,
ty $= x + t^{2}$
and $y + t x = 2 t + t^{3}$
From equation (i) and (ii) we get-
$y + t (t y - t^{2}) = 2 t + t^{3}$
$y (1 + t^{2}) - t^{3} = 2 t + t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y = \frac{2 t (1 + t^{2})}{(1 + t^{2})}$
$y = 2 t \Rightarrow y^{2} = 4 t^{2}$
Then, $x = t^{2} \Rightarrow y^{2} = 4 x$
So, the point of intersection lies on the curve $y^{2} = 4 x$

AMU-2016

Parabola

120102 The point on the parabola $y^{2} = 64 x$ which is nearest to the line $4 x + 3 y + 35 = 0$ has coordinates

1

(9, - 24)

2

(1, - 81)

3

(4, - 16)

4

(- 9, - 24)

Explanation:

A :Given, $y^{2} = 64 x$
Differentiate, w.r.t. ' $x$ '
$2 y \frac{d y}{d x} = 64$
$\frac{d y}{d x} = \frac{32}{y}$
Also,
$4 x + 3 y + 35 = 0$
$y = - \frac{4 x}{3} - \frac{35}{3}$ Hence, slope $(m) = \frac{- 4}{3} = \frac{dy}{dx}$
$\frac{- 4}{3} = \frac{32}{y}$
$y = - 24$
$∴ y^{2} = 64 x$
$x = \frac{y^{2}}{64} = \frac{(- 24)^{2}}{64}$
$x = 9$
Hence, required point is $= (9, - 24)$

WB JEE-2014

Parabola

120103 The value of $λ$ for which the curve
$(7 x + 5)^{2} + (7 y + 3)^{2} = λ^{2} (4 x + 3 y - 24)^{2}$
represents a parabola is

1

\pm \frac{6}{5}

2

\pm \frac{7}{5}

3

\pm \frac{1}{5}

4

\pm \frac{2}{5}

Explanation:

B Given,
$(7 x + 5)^{2} + (7 y + 3)^{3} = λ^{2} (4 x + 3 y - 24)^{2}$
$49 x^{2} + 25 + 70 x + 49 y^{2} + 9 + 42 y = λ^{2} (16 x^{2} + 9 y^{2} +$
$576 + 24 xy - 144 y - 192 x)$
$(49 - 16 λ)^{2} x^{2} + (49 - 9 λ^{2}) y^{2} + (70 + 192 λ^{2}) x + (42 +$
$144 λ) y - 24 λ^{2} xy + (25 - 576 λ^{2}) = 0$
$If, ax^{2} + 2 hxy + by y^{2} + 2 yx + 2 fy + c = 0$
$Then, for parabola, ab = h^{2}$
$Hence, (49 - 16 λ^{2}) (49 - 9 λ^{2}) = {(- 12 λ^{2})}^{2}$
$2401 - 441 λ^{2} - 784 λ^{2} + 144 λ^{4} = 144 λ^{4}$
$2401 - 1225 λ^{2} = 0$
$λ^{2} = \frac{2401}{1225}$
$λ = \pm \frac{7}{5}$

WB JEE-2014

Parabola

120104 A line passing through the point of intersection of $x + y = 4$ and $x - y = 2$ makes an angle $\tan^{- 1}$ $(\frac{3}{2})$ with the $X$ -axis. It intersects the parabola $y^{2} = 4 (x - 3)$ at points $(x_{1} - x_{2})$ and $(x_{2}, y_{2})$ respectively. Then, $| x_{1} - x_{2} |$ is equal to

1

\frac{16}{9}

2

\frac{32}{9}

3

\frac{40}{9}

4

\frac{80}{9}

Explanation:

B Given,
Lines: $x + y = 4 & x - y = 2$
$y = 4 - x & y = x - 2$
$∴ 4 - x = x - 2$
$2 x = 6$
$x = 3$ and $y = 1$
Now, equation of line is $(3, 1)$ and slope $(\frac{3}{4})$
$(y - 1) = \frac{3}{4} (x - 3)$
$3 x - 4 y = 5$
Also, $y^{2} = 4 (x - 3)$
${(\frac{3 x - 5}{4})}^{2} = 4 (x - 3)$
$9 x^{2} + 25 - 30 x = 64 x - 192$
$9 x^{2} - 94 x + 217 = 0$
$x = 7, \frac{31}{9}$
Hence, $| x_{1} - x_{2} | = | 7 - \frac{31}{9} |$
$| x_{1} - x_{2} | = \frac{32}{9}$

WB JEE-2016

Parabola

120100 The equation $y^{2} + 4 x + 4 y + k = 0$ represents a parabola whose latusrectum is

1 1

2 2

3 3

4 4

Explanation:

D Given,
$y^{2} + 4 x + 4 y + k = 0$
$y^{2} + 4 y + (2)^{2} - (2)^{2} + 4 x + k = 0$
$(y + 2)^{2} = - 4 x - k + 4$
$(y + 2)^{2} = - 4 (x + \frac{k}{4} - 1)$
$(y + 2)^{2} = - 4 (x + \frac{k - 4}{4})$
Now, compare with,
${(y - y_{1})}^{2} = - 4 a (x - x_{1})$ Hence, latusrectum is $4 a = 4 \times 1 = 4$

WB JEE-2012

Parabola

120101 The two lines $t y = x + t^{2}$ and $y + t x = 2 t + t^{3}$ intersect at the point lines on the curve whose equation is

1

y^{2} = 4 x

2

y^{2} = - 4 x

3

x^{2} = 4 y

4

x^{2} = - 4 y

Explanation:

A Given,
ty $= x + t^{2}$
and $y + t x = 2 t + t^{3}$
From equation (i) and (ii) we get-
$y + t (t y - t^{2}) = 2 t + t^{3}$
$y (1 + t^{2}) - t^{3} = 2 t + t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y (1 + t^{2}) = 2 t + 2 t^{3}$
$y = \frac{2 t (1 + t^{2})}{(1 + t^{2})}$
$y = 2 t \Rightarrow y^{2} = 4 t^{2}$
Then, $x = t^{2} \Rightarrow y^{2} = 4 x$
So, the point of intersection lies on the curve $y^{2} = 4 x$

AMU-2016

Parabola

120102 The point on the parabola $y^{2} = 64 x$ which is nearest to the line $4 x + 3 y + 35 = 0$ has coordinates

1

(9, - 24)

2

(1, - 81)

3

(4, - 16)

4

(- 9, - 24)

Explanation:

A :Given, $y^{2} = 64 x$
Differentiate, w.r.t. ' $x$ '
$2 y \frac{d y}{d x} = 64$
$\frac{d y}{d x} = \frac{32}{y}$
Also,
$4 x + 3 y + 35 = 0$
$y = - \frac{4 x}{3} - \frac{35}{3}$ Hence, slope $(m) = \frac{- 4}{3} = \frac{dy}{dx}$
$\frac{- 4}{3} = \frac{32}{y}$
$y = - 24$
$∴ y^{2} = 64 x$
$x = \frac{y^{2}}{64} = \frac{(- 24)^{2}}{64}$
$x = 9$
Hence, required point is $= (9, - 24)$

WB JEE-2014

Parabola

120103 The value of $λ$ for which the curve
$(7 x + 5)^{2} + (7 y + 3)^{2} = λ^{2} (4 x + 3 y - 24)^{2}$
represents a parabola is

1

\pm \frac{6}{5}

2

\pm \frac{7}{5}

3

\pm \frac{1}{5}

4

\pm \frac{2}{5}

Explanation:

B Given,
$(7 x + 5)^{2} + (7 y + 3)^{3} = λ^{2} (4 x + 3 y - 24)^{2}$
$49 x^{2} + 25 + 70 x + 49 y^{2} + 9 + 42 y = λ^{2} (16 x^{2} + 9 y^{2} +$
$576 + 24 xy - 144 y - 192 x)$
$(49 - 16 λ)^{2} x^{2} + (49 - 9 λ^{2}) y^{2} + (70 + 192 λ^{2}) x + (42 +$
$144 λ) y - 24 λ^{2} xy + (25 - 576 λ^{2}) = 0$
$If, ax^{2} + 2 hxy + by y^{2} + 2 yx + 2 fy + c = 0$
$Then, for parabola, ab = h^{2}$
$Hence, (49 - 16 λ^{2}) (49 - 9 λ^{2}) = {(- 12 λ^{2})}^{2}$
$2401 - 441 λ^{2} - 784 λ^{2} + 144 λ^{4} = 144 λ^{4}$
$2401 - 1225 λ^{2} = 0$
$λ^{2} = \frac{2401}{1225}$
$λ = \pm \frac{7}{5}$

WB JEE-2014

Parabola

120104 A line passing through the point of intersection of $x + y = 4$ and $x - y = 2$ makes an angle $\tan^{- 1}$ $(\frac{3}{2})$ with the $X$ -axis. It intersects the parabola $y^{2} = 4 (x - 3)$ at points $(x_{1} - x_{2})$ and $(x_{2}, y_{2})$ respectively. Then, $| x_{1} - x_{2} |$ is equal to

1

\frac{16}{9}

2

\frac{32}{9}

3

\frac{40}{9}

4

\frac{80}{9}

Explanation:

B Given,
Lines: $x + y = 4 & x - y = 2$
$y = 4 - x & y = x - 2$
$∴ 4 - x = x - 2$
$2 x = 6$
$x = 3$ and $y = 1$
Now, equation of line is $(3, 1)$ and slope $(\frac{3}{4})$
$(y - 1) = \frac{3}{4} (x - 3)$
$3 x - 4 y = 5$
Also, $y^{2} = 4 (x - 3)$
${(\frac{3 x - 5}{4})}^{2} = 4 (x - 3)$
$9 x^{2} + 25 - 30 x = 64 x - 192$
$9 x^{2} - 94 x + 217 = 0$
$x = 7, \frac{31}{9}$
Hence, $| x_{1} - x_{2} | = | 7 - \frac{31}{9} |$
$| x_{1} - x_{2} | = \frac{32}{9}$

WB JEE-2016

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Question Bank Series

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