Let frac{x^2}{f(a^2 + 7a + 3)} + frac{y^2}{f( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For an ellipse with the major axis along the y-axis, the denominator of the $y^2$ term must be greater than the denominator of the $x^2$ term, and

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) For an ellipse with the major axis along the y-axis, the denominator of the $y^2$ term must be greater than the denominator of the $x^2$ term, and both must be positive: $f(3a+15) > f(a^2+7a+3) > 0$.Since $f$ is a strictly decreasing function, $f(x_1) > f(x_2) \implies x_1 Therefore, $3a+15 Rearranging the inequality gives $a^2 + 4a - 12 > 0$.Factoring the quadratic expression, we get $(a+6)(a-2) > 0$.This inequality holds when $a \in (-\infty, -6) \cup (2, \infty)$.The set of values for $a$ is $R - [-6, 2]$.Comparing this with $R - [\alpha, \beta]$, we get $\alpha = -6$ and $\beta = 2$.Thus, $\alpha^2 + \beta^2 = (-6)^2 + (2)^2 = 36 + 4 = 40$.

Let $\frac{x^2}{f(a^2 + 7a + 3)} + \frac{y^2}{f(3a + 15)} = 1$ represent an ellipse with major axis along the y-axis, where $f$ is a strictly decreasing positive function on $R$. If the set of all possible values of $a$ is $R - [\alpha, \beta]$, then $\alpha^2 + \beta^2$ is equal to:

Similar Questions

Let for two distinct values of $p$ the lines $y=x+p$ touch the ellipse $E: \frac{x^2}{16} + \frac{y^2}{9} = 1$ at the points $A$ and $B$. Let the line $y = x$ intersect $E$ at the points $C$ and $D$. Then the area of the quadrilateral $ABCD$ is equal to

Tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at all the ends of its latus rectum. The area of the quadrilateral so formed (in sq. units) is

$a$ and $b$ are the semi-major and semi-minor axes of an ellipse whose axes are along the coordinate axes. If its latus rectum is of length $4$ units and the distance between its foci is $4 \sqrt{2}$,then $a^2+b^2=$

If the origin is the centre,the $X$-axis is the major axis,and $\sqrt{\frac{2}{5}}$ is the eccentricity of an ellipse which passes through $(-3, 1)$,then the equation of that ellipse is:

Find the equation of the ellipse,whose length of the major axis is $20$ and foci are $(0, \pm 5)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module