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The most general form of the equation of a conic is given by $a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 f y equals 0$ . Let x and y-axes be respectively tangent and normal to the conic given by equation ((a) at the origin. Then clearly c = 0. Also by putting y = 0, we get $a x to the power of 2 end exponent plus 2 g x$ = 0. Both roots of the equation are zero. Therefore g = 0.
Hence the most general form of the equation of such a conic is given by $blank a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 f y equals 0$ .
Let foci of the conic represented by the equation $blank a x to the power of 2 end exponent plus 2 left parenthesis a plus 2 right parenthesis x y plus a y to the power of 2 end exponent plus 2 f y equals 0$ , where $a less than negative 1$ and $f not equal to 0$ be $F subscript 1 end subscript left parenthesis 2 comma negative 3 right parenthesis blank$ and $blank F subscript 2 end subscript left parenthesis 8 comma negative 5 right parenthesis$ . Then the feet of perpendiculars from F₁ and F₂ upon x-axis lie on the curve

$x^{2} + y^{2} - 10 x + 8 y + 16 = 0$
$x^{2} + y^{2} - 10 x + 8 y = 0$
$x^{2} + y^{2} = 25$
$2 x^{2} + 8 x y - 10 y^{2} + 16 = 0$

The correct answer is: $x to the power of 2 end exponent plus y to the power of 2 end exponent minus 10 x plus 8 y plus 16 equals 0$

For $a less than negative 1$ , the equation represents an ellipse with centre at (5,-4). Also product of perpendiculars from foci upon any tangent $equals B to the power of 2 end exponent rightwards double arrow B to the power of 2 end exponent equals 15$
Also distance between foci = 2Ae = $square root of 36 plus 4 end root equals 2 square root of 10$
$A to the power of 2 end exponent e to the power of 2 end exponent equals 10 rightwards double arrow A to the power of 2 end exponent minus B to the power of 2 end exponent equals 10 rightwards double arrow A to the power of 2 end exponent equals 25$ (A and B are semi axes)
The feet of perpendicular lie on the auxiliary circle $left parenthesis x minus 5 right parenthesis to the power of 2 end exponent plus left parenthesis y plus 4 right parenthesis to the power of 2 end exponent equals 25$

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Consider the ellipse whose equation is $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ For certain pair of diameters of the above ellipse, the product of their slopes is equal to $negative fraction numerator b to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction$
The equations of two diameters are $y equals x$ and $3 y equals negative 2 x$ satisfying the condition mentioned above. Then the eccentricity of the ellipse is

Consider the ellipse whose equation is $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ For certain pair of diameters of the above ellipse, the product of their slopes is equal to $negative fraction numerator b to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction$
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The smallest possible value of S $equals a subscript 1 end subscript times a subscript 2 end subscript times a subscript 3 end subscript plus b subscript 1 end subscript times b subscript 2 end subscript times b subscript 3 end subscript plus c subscript 1 end subscript times c subscript 2 end subscript times c subscript 3 end subscript text end text$ where $text end text a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript comma b subscript 1 end subscript comma b subscript 2 end subscript comma b subscript 3 end subscript comma c subscript 1 end subscript comma c subscript 2 end subscript comma c subscript 3 end subscript$ is a permutation of the number 1, 2, 3, 4, 5, 6, 7, 8, and 9 is

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Statement 1 : The second degree equation $4 x to the power of 2 end exponent plus 9 y to the power of 2 end exponent minus 24 x plus 36 y minus 72 equals 0$ represents an ellipse Because
Statement 2 : $a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 g x plus 2 f y plus c equals 0$ represents an ellipse if $capital delta equals a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent not equal to 0$ and $h to the power of 2 end exponent greater than a b$

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Statement 2 : $a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 g x plus 2 f y plus c equals 0$ represents an ellipse if $capital delta equals a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent not equal to 0$ and $h to the power of 2 end exponent greater than a b$

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