If the line $x\cos \alpha +y\sin \alpha =p$ be normal to the ellipse $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ , then

Question

Accepted Answer

${{p}^{2}}({{a}^{2}}{{\sec }^{2}}\alpha +{{b}^{2}}	ext{cose}{{	ext{c}}^{2}}\alpha )={{({{a}^{2}}-{{b}^{2}})}^{2}}$

Answer

${{p}^{2}}({{a}^{2}}{{\cos }^{2}}\alpha +{{b}^{2}}{{\sin }^{2}}\alpha )={{a}^{2}}-{{b}^{2}}$

Answer

${{p}^{2}}({{a}^{2}}{{\cos }^{2}}\alpha +{{b}^{2}}{{\sin }^{2}}\alpha )={{({{a}^{2}}-{{b}^{2}})}^{2}}$

Answer

${{p}^{2}}({{a}^{2}}{{\sec }^{2}}\alpha +{{b}^{2}}	ext{cose}{{	ext{c}}^{2}}\alpha )={{a}^{2}}-{{b}^{2}}$

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