If tangent and normal are drawn at the end 'P' of a latus rectum in first quadrant of hyperbola $$x24$$−$$y212=1$$, meet the transverse axis at T and G respectively, then the area of the triangle PTG is

Question

If tangent and normal are drawn at the end 'P' of a latus rectum in first quadrant of  hyperbola $$x24$$−$$y212=1$$, meet the transverse axis at T and G respectively, then the area of  the triangle PTG is

Infinity Learn · Accepted Answer

Given hyperbola is $$x24$$−$$y212=1$$ Here $$a2=4$$,$$b2=12$$ Eccentricity, $$e=a2+b2a2=2P=$$ end point of latusrectum ae,$$b2a=(4$$,$$6)$$ Equation of tangent at P is $$S1=0$$⇒$$2x$$−$$y=2$$ it cuts x axis ⇒$$y=o$$ ⇒$$x=1$$ It meets tranverse axis at T $$1$$,$$0equation$$ of normal at P $$4$$,$$6$$ is $$a2$$ $$xx1+b2$$ $$yy1=a2+b2$$⇒$$4x4+12$$ $$y6=16$$    ⇒$$x+2y=16$$    it cuts  $$x-axis$$ then $$y=0$$ ,hence $$x=16$$      ⇒  $$G=16$$,$$0$$∴ Area of Δ$$PTG=12x1$$−$$x2x1$$−$$x3y1$$−$$y2y1$$−$$y3$$ $$=124$$−$$14$$−$$166$$−$$06$$−$$0=45$$ sq.units

$If tangent and normal are drawn at the end 'P' of a latus rectum in first quadrant of$
$hyperbola \frac{x^{2}}{4} - \frac{y^{2}}{12} = 1, meet the transverse axis at T and G respectively, then the area of$ $the triangle P T G is$

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If tangent and normal are drawn at the end 'P' of a latus rectum in first quadrant of hyperbola x24−y212=1, meet the transverse axis at T and G respectively, then the area of the triangle PTG is

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$If tangent and normal are drawn at the end 'P' of a latus rectum in first quadrant of$
$hyperbola \frac{x^{2}}{4} - \frac{y^{2}}{12} = 1, meet the transverse axis at T and G respectively, then the area of$ $the triangle P T G is$