Chords of the curve 4x^2 + y^2 - x + 4y = 0 w | vedclass

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Question

(A) The equation of the curve is $4x^2 + y^2 - x + 4y = 0$. Let the equation of the chord be $y = mx + c$,which implies $\frac{y - mx}{c} = 1$. Homoge

Vedclass · Accepted Answer

$\left( \frac{1}{5}, - \frac{4}{5} \right)$

Vedclass · Answer

(A) The equation of the curve is $4x^2 + y^2 - x + 4y = 0$. Let the equation of the chord be $y = mx + c$,which implies $\frac{y - mx}{c} = 1$. Homogenizing the curve equation using the line equation: $4x^2 + y^2 - x\left( \frac{y - mx}{c} \right) + 4y\left( \frac{y - mx}{c} \right) = 0$. Multiplying by $c$: $4cx^2 + cy^2 - xy + mx^2 + 4y^2 - 4mxy = 0$. $(4c + m)x^2 - (1 + 4m)xy + (c + 4)y^2 = 0$. Since the chord subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(4c + m) + (c + 4) = 0$. $5c + m + 4 = 0$,which gives $c = -\frac{m}{5} - \frac{4}{5}$. Substituting $c$ back into the line equation $y = mx + c$: $y = mx - \frac{m}{5} - \frac{4}{5}$. $y + \frac{4}{5} = m\left( x - \frac{1}{5} \right)$. This equation represents a family of lines passing through the fixed point $\left( \frac{1}{5}, -\frac{4}{5} \right)$. Wait,re-evaluating the sign: $5c + m + 4 = 0 \implies c = -\frac{m}{5} - \frac{4}{5}$. $y = mx - \frac{m}{5} - \frac{4}{5} \implies y + \frac{4}{5} = m(x - \frac{1}{5})$. This passes through $\left( \frac{1}{5}, -\frac{4}{5} \right)$.

Chords of the curve $4x^2 + y^2 - x + 4y = 0$ which subtend a right angle at the origin pass through a fixed point whose coordinates are:

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