Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3,\,0),\,\, a=4$
Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3,\,0),\,\, a=4$
Foci $(\pm 3,0), a=4$
since the foci are on the $x-$ axis, the major axis is along the $x-$ axis.
Therefore, the equation of the ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ where a is the semimajor axis.
Accordingly, $c=3$ and $a=4$
It is known that $a^{2}=b^{2}+c^{2}$
$\therefore 4^{2}=b^{2}+3^{2}$
$\Rightarrow 16=b^{2}+9$
$\Rightarrow b^{2}=16-9=7$
Thus, the equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$
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Statement $-1$ : If two tangents are drawn to an ellipse from a single point and if they are perpendicular to each other, then locus of that point is always a circle
Statement $-2$ : For an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , locus of that point from which two perpendicular tangents are drawn, is $x^2 + y^2 = (a + b)^2$ .
Statement $-1$ : If two tangents are drawn to an ellipse from a single point and if they are perpendicular to each other, then locus of that point is always a circle
Statement $-2$ : For an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , locus of that point from which two perpendicular tangents are drawn, is $x^2 + y^2 = (a + b)^2$ .