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(D) Let the equation of the line be $y = m(x-1)$,which implies $\frac{mx-y}{m} = 1$.Given the curve equation $2x^2 + 5y^2 - 7x = 0$ ... $(i)$.Using th

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(D) Let the equation of the line be $y = m(x-1)$,which implies $\frac{mx-y}{m} = 1$.Given the curve equation $2x^2 + 5y^2 - 7x = 0$ ... $(i)$.Using the method of homogenization to find the pair of lines passing through the origin and points $A$ and $B$:$2x^2 + 5y^2 - 7x(1) = 0$Substituting $1 = \frac{mx-y}{m}$:$2x^2 + 5y^2 - 7x\left(\frac{mx-y}{m}\right) = 0$Multiplying by $m$:$2mx^2 + 5my^2 - 7mx^2 + 7xy = 0$$-5mx^2 + 7xy + 5my^2 = 0$.This is a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$,where $a = -5m$,$2h = 7$,and $b = 5m$.For the lines to be perpendicular,the condition is $a + b = 0$.Here,$a + b = -5m + 5m = 0$.Since the sum of the coefficients of $x^2$ and $y^2$ is zero,the lines are perpendicular.Therefore,the angle subtended by the line segment $AB$ at the origin is $90^\circ$.

$A$ straight line passing through the point $(1,0)$ and not parallel to the $x$-axis intersects the curve $2x^2+5y^2-7x=0$ at two points $A$ and $B$. The angle subtended by the line segment $AB$ at the origin is (in $^\circ$)

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