A) parallel
B) perpendicular
C) inclined at an angle\[\frac{n(n+1)d}{2n+1}\]
D) None of these
Correct Answer: A
Solution :
At the point where the given curve crosses X-axis, we have \[y=0\]\[\Rightarrow \]\[a{{x}^{2}}=1\] [putting y = 0 in\[a{{x}^{2}}+2hyx+b{{y}^{2}}=1\]] \[\Rightarrow \]\[x=\pm \frac{1}{\sqrt{a}}\] Thus, the given curve cuts X-axis at\[P\left( \frac{1}{\sqrt{a}},0 \right)\] and\[Q\left( -\frac{1}{\sqrt{a}},0 \right).\] Now, \[a{{x}^{2}}+2hxy+b{{y}^{2}}=1\] On differentiating both sides w.r.t. x, we get \[2ax+2h\left( x\frac{dy}{dx}+y \right)+2by\frac{dy}{dx}=0\] \[\Rightarrow \]\[\frac{dy}{dx}=-\frac{ax+hy}{hx+by}\] \[\Rightarrow \]\[{{\left[ \frac{dy}{dx} \right]}_{P}}=-\frac{a}{h}\]and \[{{\left[ \frac{dy}{dx} \right]}_{Q}}=-\frac{a}{h}\] \[\Rightarrow \]\[{{\left[ \frac{dy}{dx} \right]}_{p}}={{\left[ \frac{dy}{dx} \right]}_{Q}}\] Hence, the tangents are parallel.
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