A) \[{{p}^{2}}({{a}^{2}}{{\cos }^{2}}\alpha +{{b}^{2}}{{\sin }^{2}}\alpha )={{a}^{2}}-{{b}^{2}}\]
B) \[{{p}^{2}}({{a}^{2}}{{\cos }^{2}}\alpha +{{b}^{2}}{{\sin }^{2}}\alpha )={{({{a}^{2}}-{{b}^{2}})}^{2}}\]
C) \[{{p}^{2}}({{a}^{2}}{{\sec }^{2}}\alpha +{{b}^{2}}\text{cose}{{\text{c}}^{2}}\alpha )={{a}^{2}}-{{b}^{2}}\]
D) \[{{p}^{2}}({{a}^{2}}{{\sec }^{2}}\alpha +{{b}^{2}}\text{cose}{{\text{c}}^{2}}\alpha )={{({{a}^{2}}-{{b}^{2}})}^{2}}\]
Correct Answer: D
Solution :
The equation of any normal to \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[ax\,\,\sec \varphi -by\,\,\text{cosec}\,\varphi ={{a}^{2}}-{{b}^{2}}\] .....(i) The straight line \[x\cos \alpha +y\sin \alpha =p\] will be a normal to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] If (i) and \[x\cos \alpha +y\sin \alpha =p\] represent the same line \[\frac{a\sec \varphi }{\cos \alpha }=\frac{-b\,\text{cosec}\varphi }{\sin \alpha }=\frac{{{a}^{2}}-{{b}^{2}}}{p}\] \[\Rightarrow \,\cos \varphi \,=\frac{ap}{({{a}^{2}}-{{b}^{2}})\cos \alpha },\,\]\[\sin \varphi =\frac{-bp}{({{a}^{2}}-{{b}^{2}})\sin \alpha }\] \[\because \]\[{{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi =1\] \[\Rightarrow \,\frac{{{b}^{2}}{{p}^{2}}}{{{({{a}^{2}}-{{b}^{2}})}^{2}}{{\sin }^{2}}\alpha }+\frac{{{a}^{2}}{{p}^{2}}}{{{({{a}^{2}}-{{b}^{2}})}^{2}}{{\cos }^{2}}\alpha }=1\] \[\Rightarrow \]\[{{p}^{2}}({{b}^{2}}\text{cose}{{\text{c}}^{\text{2}}}\,\alpha +{{a}^{2}}{{\sec }^{2}}\alpha )={{({{a}^{2}}-{{b}^{2}})}^{2}}\].
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