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Maths NCERT Exemplar Solutions Class 12th Chapter Eight Maths Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Conic Sections

If the equation of the parabola, whose vertex is at (5, 4) and the directrix is 3x + y – 29 = 0, is $x^{2} + a y^{2} + b x y + c x + d y + k = 0,$ then a + b + c + d + k is equal to

Option 1 -

575

Option 2 -

-575

Option 3 -

576

Option 4 -

-576

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

Given vertex is (5, 4) and directrix 3x + y – 29 = 0

Let foot of perpendicular of (5, 4) on directrix be (x₁, y₁)

$\frac{x_{1} - 5}{3} = \frac{y_{1} - 4}{1} = \frac{- (- 10)}{10}$

$∴ (x_{1}, y_{1}) = (8, 5)$

So, focus of parabola will be S = (2, 3)

Let P(x, y) be any point on parabola, then

$\Rightarrow {(x - 2)}^{2} + {(y - 3)}^{2} = \frac{{(3 x + y - 29)}^{2}}{10}$

$\Rightarrow x^{2} + 9 y^{2} - 6 x y + 134 x - 2 y - 711 = 0$

And given parabola

$x^{2} + a y^{2} + b x y + c x + d y + k = 0$

$∴ a = 9, b = - 6, c = 134, d = - 2, k = - 711$

$∴ a + b + c + d + k = 576$

Given vertex is  (5, 4) and directrix 3x + y – 29 = 0Let foot of perpendicular of (5, 4) on directrix be (x1, y1) <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>x</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <mn>5</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <msub> <mrow> <mi>y</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <mn>4</mn> </mrow> <mrow> <mn>1</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mo>−</mo> <mrow> <mo>(</mo> <mrow> <mo>−</mo> <mn>1</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </mfrac> </mrow> </math>                <math> <mrow> <mo>∴</mo> <mrow> <mo>(</mo> <mrow> <msub> <mrow> <mi>x</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>,</mo> <msub> <mrow> <mi>y</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mrow> <mo>(</mo> <mrow> <mn>8</mn> <mo>,</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math>               So, focus of parabola will be S = (2, 3)Let P(x, y) be any point on parabola, then <math> <mrow> <mo>⇒</mo> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>y</mi> <mo>−</mo> <mn>3</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>3</mn> <mi>x</mi> <mo>+</mo> <mi>y</mi> <mo>−</mo> <mn>2</mn> <mn>9</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </mfrac> </mrow> </math>                <math> <mrow> <mo>⇒</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>9</mn> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>6</mn> <mi>x</mi> <mi>y</mi> <mo>+</mo> <mn>1</mn> <mn>3</mn> <mn>4</mn> <mi>x</mi> <mo>−</mo> <mn>2</mn> <mi>y</mi> <mo>−</mo> <mn>7</mn> <mn>1</mn> <mn>1</mn> <mo>=</mo> <mn>0</mn> </mrow> </math>               And given parabola <math> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mi>a</mi> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mi>b</mi> <mi>x</mi> <mi>y</mi> <mo>+</mo> <mi>c</mi> <mi>x</mi> <mo>+</mo> <mi>d</mi> <mi>y</mi> <mo>+</mo> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> </math>                <math> <mrow> <mo>∴</mo> <mi>a</mi> <mo>=</mo> <mn>9</mn> <mo>,</mo> <mi>b</mi> <mo>=</mo> <mo>−</mo> <mn>6</mn> <mo>,</mo> <mi>c</mi> <mo>=</mo> <mn>1</mn> <mn>3</mn> <mn>4</mn> <mo>,</mo> <mi>d</mi> <mo>=</mo> <mo>−</mo> <mn>2</mn> <mo>,</mo> <mi>k</mi> <mo>=</mo> <mo>−</mo> <mn>7</mn> <mn>1</mn> <mn>1</mn> </mrow> </math>                <math> <mrow> <mo>∴</mo> <mi>a</mi> <mo>+</mo> <mi>b</mi> <mo>+</mo> <mi>c</mi> <mo>+</mo> <mi>d</mi> <mo>+</mo> <mi>k</mi> <mo>=</mo> <mn>5</mn> <mn>7</mn> <mn>6</mn> </mrow> </math>

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If the equation of the parabola, whose vertex is at (5, 4) and the directrix is 3x + y – 29 = 0, is $x^{2} + a y^{2} + b x y + c x + d y + k = 0,$ then a + b + c + d + k is equal to

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If the equation of the parabola, whose vertex is at (5, 4) and the directrix is 3x + y – 29 = 0, is x 2 + a y 2 + b x y + c x + d y + k = 0 , then a + b + c + d + k is equal to

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If the equation of the parabola, whose vertex is at (5, 4) and the directrix is 3x + y – 29 = 0, is $x^{2} + a y^{2} + b x y + c x + d y + k = 0,$ then a + b + c + d + k is equal to