A vertical line passing through the point $ \left( h,0 \right) $ intersects the ellipse $ \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{3}=1 $ at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If $ \Delta \left( h \right) $ = area of the triangle PQR, $ {{\Delta }_{1}}=\displaystyle \max_{\dfrac{1}{2}\le h\le 1}\left( \Delta \left( h \right) \right) $ and $ {{\Delta }_{2}}=\displaystyle \min_{\dfrac{1}{2}\le h\le 1}\left( \Delta \left( h \right) \right) $ , then find the value of $ \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}} $ ?
(a) 3
(b) 6
(c) 9
(d) 12
Answer
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Hint: We start solving the problem by drawing the figure representing the given information. We then find the equation of the vertical line by making use of the fact that the equation of the vertical line passing through the point $ \left( a,0 \right) $ is $ x=a $ . We then find the points of intersection of this line P and Q with the ellipse. We then find the equation of tangents at the points of intersection by making use of the fact that the equation of the tangent to the ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ at point $ \left( {{x}_{1}},{{y}_{1}} \right) $ on the ellipse is $ \dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1 $ . We then find the point of intersection of tangents R and then find the area of the triangle PQR. We then differentiate it and check its values to check whether it is decreasing or increasing function. We then find the value of $ {{\Delta }_{1}} $ , $ {{\Delta }_{2}} $ and substitute those values in $ \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}} $ to get the required answer.
Complete step by step answer:
According to the problem, we are given that a vertical line passing through the point $ \left( h,0 \right) $ intersects the ellipse $ \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{3}=1 $ at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. We need to find the value of $ \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}} $ if $ \Delta \left( h \right) $ = area of the triangle PQR, $ {{\Delta }_{1}}=\displaystyle \max_{\dfrac{1}{2}\le h\le 1}\left( \Delta \left( h \right) \right) $ and $ {{\Delta }_{2}}=\displaystyle \min_{\dfrac{1}{2}\le h\le 1}\left( \Delta \left( h \right) \right) $ .
Let us draw the figure representing the given information.
We know that the equation of the vertical line passing through the point $ \left( a,0 \right) $ is $ x=a $ .
So, the vertical line given in the problem is $ x=h $ . Now, let us find the point of intersection of this line with ellipse $ \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{3}=1 $ .
So, we have $ \dfrac{{{h}^{2}}}{4}+\dfrac{{{y}^{2}}}{3}=1 $ .
$ \Rightarrow \dfrac{{{y}^{2}}}{3}=1-\dfrac{{{h}^{2}}}{4} $ .
$ \Rightarrow {{y}^{2}}=\dfrac{3\left( 4-{{h}^{2}} \right)}{4} $ .
$ \Rightarrow y=\pm \dfrac{\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} $ .
So, we get the point P and Q as $ P\left( h,\dfrac{\sqrt{3}\sqrt{\left( 4-{{h}^{2}} \right)}}{2} \right) $ and $ Q\left( h,\dfrac{-\sqrt{3}\sqrt{\left( 4-{{h}^{2}} \right)}}{2} \right) $ .
We know that the equation of the tangent to the ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ at point $ \left( {{x}_{1}},{{y}_{1}} \right) $ on the ellipse is $ \dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1 $ .
So, the equation of the tangent at point $ P\left( h,\dfrac{\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right) $ is \[\dfrac{xh}{4}+\dfrac{y\left( \dfrac{\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right)}{3}=1\].
$ \Rightarrow \dfrac{xh}{4}+\dfrac{y\left( \sqrt{4-{{h}^{2}}} \right)}{2\sqrt{3}}=1 $ .
$ \Rightarrow \dfrac{\left( h\sqrt{3} \right)x+2y\left( \sqrt{4-{{h}^{2}}} \right)}{4\sqrt{3}}=1 $ .
$ \Rightarrow \left( h\sqrt{3} \right)x+2y\left( \sqrt{4-{{h}^{2}}} \right)=4\sqrt{3} $ ---(1).
So, the equation of the tangent at point $ Q\left( h,\dfrac{-\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right) $ is \[\dfrac{xh}{4}+\dfrac{y\left( \dfrac{-\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right)}{3}=1\].
$ \Rightarrow \dfrac{xh}{4}-\dfrac{y\left( \sqrt{4-{{h}^{2}}} \right)}{2\sqrt{3}}=1 $ .
$ \Rightarrow \dfrac{\left( h\sqrt{3} \right)x-2y\left( \sqrt{4-{{h}^{2}}} \right)}{4\sqrt{3}}=1 $ .
$ \Rightarrow \left( h\sqrt{3} \right)x-2y\left( \sqrt{4-{{h}^{2}}} \right)=4\sqrt{3} $ ---(2).
Let us add equations (1) and (2). So, we get
$ \Rightarrow \left( 2h\sqrt{3} \right)x=8\sqrt{3} $ .
$ \Rightarrow x=\dfrac{4}{h} $ ---(3).
Let us substitute equation (3) in equation (2).
$ \Rightarrow \left( h\sqrt{3} \right)\left( \dfrac{4}{h} \right)-2y\left( \sqrt{4-{{h}^{2}}} \right)=4\sqrt{3} $ .
$ \Rightarrow 4\sqrt{3}-2y\left( \sqrt{4-{{h}^{2}}} \right)=4\sqrt{3} $ .
$ \Rightarrow 2y\left( \sqrt{4-{{h}^{2}}} \right)=0 $ .
$ \Rightarrow y=0 $ .
So, we got the point R as $ R\left( \dfrac{4}{h},0 \right) $ .
Now, let us find the area of the triangle PQR.
From the triangle we can see that $ \Delta \left( h \right)=\dfrac{1}{2}\times RA\times PQ $ ---(4).
Let us find the values of RA and PQ.
So, we have $ RA=\sqrt{{{\left( \dfrac{4}{h}-h \right)}^{2}}+{{\left( 0-0 \right)}^{2}}} $ .
$ \Rightarrow RA=\dfrac{4}{h}-h $ ---(5).
Now, we have $ PQ=\sqrt{{{\left( h-h \right)}^{2}}+{{\left( \left( \dfrac{-\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right)-\left( \dfrac{\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right) \right)}^{2}}} $ .
$ \Rightarrow PQ=\sqrt{0+{{\left( \dfrac{-2\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right)}^{2}}} $ .
$ \Rightarrow PQ=\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right) $ ---(6).
Let us substitute equations (5) and (6) in equation (4).
$ \Rightarrow \Delta \left( h \right)=\dfrac{1}{2}\times \left( \dfrac{4}{h}-h \right)\times \sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right) $ .
$ \Rightarrow \Delta \left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{4-{{h}^{2}}}{h} \right)\times \left( \sqrt{4-{{h}^{2}}} \right) $ .
$ \Rightarrow \Delta \left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}}{h} \right) $ ---(7).
Now, let us differentiate equation (7) on both sides.
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{h\dfrac{d}{dh}{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}-{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}\dfrac{d}{dh}\left( h \right)}{{{h}^{2}}} \right) $ .
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{h\left( \dfrac{3}{2} \right){{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}\left( -2h \right)-{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}\left( 1 \right)}{{{h}^{2}}} \right) $ .
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{\left( -3{{h}^{2}} \right){{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}-{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}}{{{h}^{2}}} \right) $ .
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}\left( -3{{h}^{2}}-4+{{h}^{2}} \right)}{{{h}^{2}}} \right) $ .
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}\left( -2{{h}^{2}}-4 \right)}{{{h}^{2}}} \right) $ .
We know that $ {{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}>0 $ , $ {{h}^{2}}>0 $ and $ -2{{h}^{2}}-4<0 $ . This tells us that $ {{\Delta }^{'}}\left( h \right)<0 $ , which means that $ \Delta \left( h \right) $ is a decreasing function.
So, we get the maximum of $ \Delta \left( h \right) $ at $ h=\dfrac{1}{2} $ and minimum of $ \Delta \left( h \right) $ at $ h=1 $ .
So, we get $ {{\Delta }_{1}}=\Delta \left( \dfrac{1}{2} \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{\left( \dfrac{1}{2} \right)}^{2}} \right)}^{\dfrac{3}{2}}}}{\left( \dfrac{1}{2} \right)} \right) $ .
$ \Rightarrow {{\Delta }_{1}}=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-\left( \dfrac{1}{4} \right) \right)}^{\dfrac{3}{2}}}}{\left( \dfrac{1}{2} \right)} \right) $ .
$ \Rightarrow {{\Delta }_{1}}=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( \dfrac{15}{4} \right)}^{\dfrac{3}{2}}}}{\left( \dfrac{1}{2} \right)} \right) $ .
$ \Rightarrow {{\Delta }_{1}}=\dfrac{45\sqrt{5}}{8} $ ---(8).
Now, we $ {{\Delta }_{2}}=\Delta \left( 1 \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{\left( 1 \right)}^{2}} \right)}^{\dfrac{3}{2}}}}{\left( 1 \right)} \right) $ .
$ \Rightarrow {{\Delta }_{2}}=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-1 \right)}^{\dfrac{3}{2}}}}{\left( 1 \right)} \right) $ .
$ \Rightarrow {{\Delta }_{2}}=\dfrac{\sqrt{3}}{2}\times \left( {{\left( 3 \right)}^{\dfrac{3}{2}}} \right) $ .
$ \Rightarrow {{\Delta }_{2}}=\dfrac{9}{2} $ ---(9).
Now, let us find the value of $ \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}} $ .
$ \Rightarrow \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}}=\dfrac{8}{\sqrt{5}}\left( \dfrac{45\sqrt{5}}{8} \right)-8\left( \dfrac{9}{2} \right) $ .
$ \Rightarrow \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}}=45-36 $ .
$ \Rightarrow \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}}=9 $ .
$ \therefore $ The correct option for the given problem is (c).
Note:
We can see that the given problem contains a huge amount of calculations, so we need to perform each step carefully to avoid confusion and calculation mistakes. We have assumed that the value of h lies in the interval $ \left( -2,2 \right) $ to solve the problem, otherwise, the solution for the given problem will be different. Similarly, we can expect problems to find the value of $ \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}} $ if R is the point of intersection of normal at points P and Q.
Complete step by step answer:
According to the problem, we are given that a vertical line passing through the point $ \left( h,0 \right) $ intersects the ellipse $ \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{3}=1 $ at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. We need to find the value of $ \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}} $ if $ \Delta \left( h \right) $ = area of the triangle PQR, $ {{\Delta }_{1}}=\displaystyle \max_{\dfrac{1}{2}\le h\le 1}\left( \Delta \left( h \right) \right) $ and $ {{\Delta }_{2}}=\displaystyle \min_{\dfrac{1}{2}\le h\le 1}\left( \Delta \left( h \right) \right) $ .
Let us draw the figure representing the given information.
We know that the equation of the vertical line passing through the point $ \left( a,0 \right) $ is $ x=a $ .
So, the vertical line given in the problem is $ x=h $ . Now, let us find the point of intersection of this line with ellipse $ \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{3}=1 $ .
So, we have $ \dfrac{{{h}^{2}}}{4}+\dfrac{{{y}^{2}}}{3}=1 $ .
$ \Rightarrow \dfrac{{{y}^{2}}}{3}=1-\dfrac{{{h}^{2}}}{4} $ .
$ \Rightarrow {{y}^{2}}=\dfrac{3\left( 4-{{h}^{2}} \right)}{4} $ .
$ \Rightarrow y=\pm \dfrac{\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} $ .
So, we get the point P and Q as $ P\left( h,\dfrac{\sqrt{3}\sqrt{\left( 4-{{h}^{2}} \right)}}{2} \right) $ and $ Q\left( h,\dfrac{-\sqrt{3}\sqrt{\left( 4-{{h}^{2}} \right)}}{2} \right) $ .
We know that the equation of the tangent to the ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ at point $ \left( {{x}_{1}},{{y}_{1}} \right) $ on the ellipse is $ \dfrac{x{{x}_{1}}}{{{a}^{2}}}+\dfrac{y{{y}_{1}}}{{{b}^{2}}}=1 $ .
So, the equation of the tangent at point $ P\left( h,\dfrac{\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right) $ is \[\dfrac{xh}{4}+\dfrac{y\left( \dfrac{\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right)}{3}=1\].
$ \Rightarrow \dfrac{xh}{4}+\dfrac{y\left( \sqrt{4-{{h}^{2}}} \right)}{2\sqrt{3}}=1 $ .
$ \Rightarrow \dfrac{\left( h\sqrt{3} \right)x+2y\left( \sqrt{4-{{h}^{2}}} \right)}{4\sqrt{3}}=1 $ .
$ \Rightarrow \left( h\sqrt{3} \right)x+2y\left( \sqrt{4-{{h}^{2}}} \right)=4\sqrt{3} $ ---(1).
So, the equation of the tangent at point $ Q\left( h,\dfrac{-\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right) $ is \[\dfrac{xh}{4}+\dfrac{y\left( \dfrac{-\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right)}{3}=1\].
$ \Rightarrow \dfrac{xh}{4}-\dfrac{y\left( \sqrt{4-{{h}^{2}}} \right)}{2\sqrt{3}}=1 $ .
$ \Rightarrow \dfrac{\left( h\sqrt{3} \right)x-2y\left( \sqrt{4-{{h}^{2}}} \right)}{4\sqrt{3}}=1 $ .
$ \Rightarrow \left( h\sqrt{3} \right)x-2y\left( \sqrt{4-{{h}^{2}}} \right)=4\sqrt{3} $ ---(2).
Let us add equations (1) and (2). So, we get
$ \Rightarrow \left( 2h\sqrt{3} \right)x=8\sqrt{3} $ .
$ \Rightarrow x=\dfrac{4}{h} $ ---(3).
Let us substitute equation (3) in equation (2).
$ \Rightarrow \left( h\sqrt{3} \right)\left( \dfrac{4}{h} \right)-2y\left( \sqrt{4-{{h}^{2}}} \right)=4\sqrt{3} $ .
$ \Rightarrow 4\sqrt{3}-2y\left( \sqrt{4-{{h}^{2}}} \right)=4\sqrt{3} $ .
$ \Rightarrow 2y\left( \sqrt{4-{{h}^{2}}} \right)=0 $ .
$ \Rightarrow y=0 $ .
So, we got the point R as $ R\left( \dfrac{4}{h},0 \right) $ .
Now, let us find the area of the triangle PQR.
From the triangle we can see that $ \Delta \left( h \right)=\dfrac{1}{2}\times RA\times PQ $ ---(4).
Let us find the values of RA and PQ.
So, we have $ RA=\sqrt{{{\left( \dfrac{4}{h}-h \right)}^{2}}+{{\left( 0-0 \right)}^{2}}} $ .
$ \Rightarrow RA=\dfrac{4}{h}-h $ ---(5).
Now, we have $ PQ=\sqrt{{{\left( h-h \right)}^{2}}+{{\left( \left( \dfrac{-\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right)-\left( \dfrac{\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right) \right)}^{2}}} $ .
$ \Rightarrow PQ=\sqrt{0+{{\left( \dfrac{-2\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right)}{2} \right)}^{2}}} $ .
$ \Rightarrow PQ=\sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right) $ ---(6).
Let us substitute equations (5) and (6) in equation (4).
$ \Rightarrow \Delta \left( h \right)=\dfrac{1}{2}\times \left( \dfrac{4}{h}-h \right)\times \sqrt{3}\left( \sqrt{4-{{h}^{2}}} \right) $ .
$ \Rightarrow \Delta \left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{4-{{h}^{2}}}{h} \right)\times \left( \sqrt{4-{{h}^{2}}} \right) $ .
$ \Rightarrow \Delta \left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}}{h} \right) $ ---(7).
Now, let us differentiate equation (7) on both sides.
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{h\dfrac{d}{dh}{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}-{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}\dfrac{d}{dh}\left( h \right)}{{{h}^{2}}} \right) $ .
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{h\left( \dfrac{3}{2} \right){{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}\left( -2h \right)-{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}\left( 1 \right)}{{{h}^{2}}} \right) $ .
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{\left( -3{{h}^{2}} \right){{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}-{{\left( 4-{{h}^{2}} \right)}^{\dfrac{3}{2}}}}{{{h}^{2}}} \right) $ .
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}\left( -3{{h}^{2}}-4+{{h}^{2}} \right)}{{{h}^{2}}} \right) $ .
$ \Rightarrow {{\Delta }^{'}}\left( h \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}\left( -2{{h}^{2}}-4 \right)}{{{h}^{2}}} \right) $ .
We know that $ {{\left( 4-{{h}^{2}} \right)}^{\dfrac{1}{2}}}>0 $ , $ {{h}^{2}}>0 $ and $ -2{{h}^{2}}-4<0 $ . This tells us that $ {{\Delta }^{'}}\left( h \right)<0 $ , which means that $ \Delta \left( h \right) $ is a decreasing function.
So, we get the maximum of $ \Delta \left( h \right) $ at $ h=\dfrac{1}{2} $ and minimum of $ \Delta \left( h \right) $ at $ h=1 $ .
So, we get $ {{\Delta }_{1}}=\Delta \left( \dfrac{1}{2} \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{\left( \dfrac{1}{2} \right)}^{2}} \right)}^{\dfrac{3}{2}}}}{\left( \dfrac{1}{2} \right)} \right) $ .
$ \Rightarrow {{\Delta }_{1}}=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-\left( \dfrac{1}{4} \right) \right)}^{\dfrac{3}{2}}}}{\left( \dfrac{1}{2} \right)} \right) $ .
$ \Rightarrow {{\Delta }_{1}}=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( \dfrac{15}{4} \right)}^{\dfrac{3}{2}}}}{\left( \dfrac{1}{2} \right)} \right) $ .
$ \Rightarrow {{\Delta }_{1}}=\dfrac{45\sqrt{5}}{8} $ ---(8).
Now, we $ {{\Delta }_{2}}=\Delta \left( 1 \right)=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-{{\left( 1 \right)}^{2}} \right)}^{\dfrac{3}{2}}}}{\left( 1 \right)} \right) $ .
$ \Rightarrow {{\Delta }_{2}}=\dfrac{\sqrt{3}}{2}\times \left( \dfrac{{{\left( 4-1 \right)}^{\dfrac{3}{2}}}}{\left( 1 \right)} \right) $ .
$ \Rightarrow {{\Delta }_{2}}=\dfrac{\sqrt{3}}{2}\times \left( {{\left( 3 \right)}^{\dfrac{3}{2}}} \right) $ .
$ \Rightarrow {{\Delta }_{2}}=\dfrac{9}{2} $ ---(9).
Now, let us find the value of $ \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}} $ .
$ \Rightarrow \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}}=\dfrac{8}{\sqrt{5}}\left( \dfrac{45\sqrt{5}}{8} \right)-8\left( \dfrac{9}{2} \right) $ .
$ \Rightarrow \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}}=45-36 $ .
$ \Rightarrow \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}}=9 $ .
$ \therefore $ The correct option for the given problem is (c).
Note:
We can see that the given problem contains a huge amount of calculations, so we need to perform each step carefully to avoid confusion and calculation mistakes. We have assumed that the value of h lies in the interval $ \left( -2,2 \right) $ to solve the problem, otherwise, the solution for the given problem will be different. Similarly, we can expect problems to find the value of $ \dfrac{8}{\sqrt{5}}{{\Delta }_{1}}-8{{\Delta }_{2}} $ if R is the point of intersection of normal at points P and Q.
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