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51 Questions around this concept.
The tangent at any point P of a given circle meets the tangent at a fixed point A in T, and T is joined to B, the other end of the diameter through A, The locus of the intersection of AP and BT is an ellipse whose eccentricity is.
The pole of the straight line with respect to ellipse
is
The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to :
Line and the Ellipse
Equation of Ellipse and Line is
Ellipse : $\quad \frac{\mathrm{x}}{\mathrm{a}^2}+\frac{\mathrm{y}}{\mathrm{b}^2}=1$
Line: $\quad y=m x+c$
After solving Eq. (i) and Eq. (ii)
$
\begin{aligned}
& \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{(\mathrm{mx}+\mathrm{c})^2}{\mathrm{~b}^2}=1 \\
& \Rightarrow \quad\left(\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2\right) \mathrm{x}^2+2 \mathrm{mca}^2 \mathrm{x}+\mathrm{c}^2 \mathrm{a}^2-\mathrm{a}^2 \mathrm{~b}^2=0
\end{aligned}
$
Above equation is quadratic in x
Now depending on the value of determinant of this equation we can have following cases
If $\mathrm{D}>0$, then we have 2 real and distinct roots which means two distinct points of intersection of the line and the ellipse.
If $D=0$, then we have 1 real and repeated root which means one point of intersection of the line and the ellipse which means that the line is tangent to the ellipse. By putting values in $\mathrm{D}=0$, we get $a^2 m^2+b^2=c^2$. This is also the condition of tangency for the line $y=m x+c$ to be tangent to the ellipse.
If $\mathrm{D}<0$, then we do not have any real root, which meansno point of intersection of the line and the ellipse.
Using $a^2 m^2+b^2=c^2$ as the condition of tangency for the line $y=m x+c$ to be tangent to the ellipse, the equation of tangent to the standard ellipse is
$y=m x \pm \sqrt{a^2 m^2+b^2}$
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