Question
The equation of a conic is $25 \mathrm{x}^{2}+16 \mathrm{y}^{2}=1600$. Then(a) its latus rectum $=25$(b) focal distances of the point $(4 \sqrt{3}, 5)$ are 7 and 13(c) line $x+4 y+4 c=0$ is a tangent to the conic if $c^{2}=\frac{281}{4}$(d) line $2 x+8 y+3=0$ is a tangent to the conic
Step 1
We can rewrite this equation in the standard form of an ellipse equation, which is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. So, the given equation becomes $\frac{x^{2}}{64}+\frac{y^{2}}{100}=1$. Show more…
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