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Tangents are drawn to the ellipse $fraction numerator x to the power of 2 end exponent over denominator 9 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 end fraction equals 1$ at the ends of the latus rectum. The are of the quadrilateral so formed is

27
$\frac{13}{2}$
$\frac{15}{4}$
45

hint Hint:

The eccentricity of an ellipse is always less than 1. i.e. e < 1. The eccentricity of an ellipse can be taken as the ratio of its distance from the focus and the distance from the directrix.
Eccentricity of ellipse (e) = $square root of 1 minus b squared over a squared end root$
Here a is the length of the semi-major axis and b is the length of the semi-minor axis.

The correct answer is: 27

Let $L L^{'}$ and $M M^{'}$ are the latus rectum.
Eccentricity of ellipse (e) = $square root of 1 minus b squared over a squared end root$
Here a is the length of the semi-major axis and b is the length of the semi-minor axis.
$e space equals space square root of 1 space minus space b squared over a squared end root F r o m comma space x squared over 9 space plus space y squared over 5 space equals 1 b squared space equals space 5 space a n d space a squared space equals space 9 e space equals space square root of 1 space minus space 5 over 9 end root e space equals space 2 over 3$
The equation of tangent at $(x_{1}, y_{1})$ is $fraction numerator x x subscript 1 over denominator 9 end fraction space plus space fraction numerator y y subscript 1 over denominator 5 end fraction space equals space 1$
The tangent passes through point $b squared over a right parenthesis$
$S o comma space fraction numerator x left parenthesis a e right parenthesis over denominator 9 end fraction space plus space fraction numerator y left parenthesis begin display style b squared over a end style right parenthesis over denominator 5 end fraction space equals 1 space space space space space space rightwards double arrow e x space plus space y space equals space a space space space space space space rightwards double arrow fraction numerator x over denominator begin display style a over e end style end fraction plus space y over a space equals space 1$
$So, the tangent cuts the x-axis at point A(a/e,0) and y-axis at point B(0,a)$
$Area of quadrilateral ABCD = 4\timesarea of △OAB$
$4 space cross times 1 half cross times a over e cross times a$
$2 space cross times a squared over e$
$2 space space cross times space fraction numerator 9 over denominator begin display style 2 over 3 end style end fraction$
$Area of quadrilateral ABCD = 27sq. units$

A tangent to the ellipse x² + 4y² = 4 meets the ellipse x² + 2y² = 6 at P and Q. The angle between the tangents at P and Q of the ellipse x² + 2y² = 6 is

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A cubic polynomial f(x) = ax³ + bx² + cx + d has a graph which touches the x-axis at 2, has another x-intercept at –1 and has y-intercept at –2 as shown. The value of, a + b + c + d equals

If a function has a tangent at a point $x = a$ , then, it means, it is the critical point of that function and the derivative of the function is zero at that point.
Perform the calculations carefully, to avoid silly mistakes.
One should know how to read a graph to solve such questions.

A cubic polynomial f(x) = ax³ + bx² + cx + d has a graph which touches the x-axis at 2, has another x-intercept at –1 and has y-intercept at –2 as shown. The value of, a + b + c + d equals

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If P (q) and Q $open parentheses fraction numerator pi over denominator 2 end fraction plus theta close parentheses$ are two points on the ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ , then locus of the mid – point of PQ is

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If CP and CD are semi – conjugate diameters of an ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 comma$ then CP² + CD² =

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The normal to the curve x = 3 cos $theta$ – cos3 $theta$ , y = 3 sin $theta$ – sin3 $theta$ at the point $theta$ = $pi$ /4 passes through the point -

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If the tangent at ‘t’ on the curve y = 8t3 –1, x = 4t2 + 3 meets the curve again at t¢ and is normal to the curve at that point, then value of t must be -

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The tangent at (t, t2 – t3) on the curve y = x2 – x3 meets the curve again at Q, then abscissa of Q must be -

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If $fraction numerator x over denominator a end fraction plus fraction numerator y over denominator b end fraction$ = 1 is a tangent to the curve x = 4t,y = $fraction numerator 4 over denominator t end fraction$ , t $element of$ R then -

Maths-General

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The line $fraction numerator x over denominator a end fraction$ + $fraction numerator y over denominator b end fraction$ = 1 touches the curve y = be-x/a at the point :

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Tangents are drawn to the ellipse at the ends of the latus rectum. The are of the quadrilateral so formed is

27 45

The correct answer is: 27

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Tangents are drawn to the ellipse $fraction numerator x to the power of 2 end exponent over denominator 9 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 end fraction equals 1$ at the ends of the latus rectum. The are of the quadrilateral so formed is

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